返回特征解读

2006 中国数学奥林匹克（CMO）

books/competition_archive/cmo/2006_cmo.pdf · HS-MATH-1024-v2.1-solution-aware

6 题10 个小问/题组
1

Day 1 January 12th · 代数

Suppose that the real numbers a1,a2,…,an\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{n}} satisfy a1+a2+…+an=0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \dots \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{n}} \htmlData{tutor-start=30,tutor-end=31}{=} \htmlData{tutor-start=32,tutor-end=33}{0}. Prove max⁡1≤i≤nai2≤n3∑i=1n−1(ai−ai+1)2.\max_{\htmlData{tutor-start=6,tutor-end=7}{1} \htmlData{tutor-start=8,tutor-end=12}{\le }\htmlData{tutor-start=12,tutor-end=13}{i} \htmlData{tutor-start=14,tutor-end=18}{\le }\htmlData{tutor-start=18,tutor-end=19}{n}} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{i}}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=35}{\le }\frac{\htmlData{tutor-start=41,tutor-end=42}{n}}{\htmlData{tutor-start=44,tutor-end=45}{3}} \sum_{\htmlData{tutor-start=53,tutor-end=54}{i}\htmlData{tutor-start=54,tutor-end=55}{=}\htmlData{tutor-start=55,tutor-end=56}{1}}^{\htmlData{tutor-start=59,tutor-end=60}{n}\htmlData{tutor-start=60,tutor-end=61}{-}\htmlData{tutor-start=61,tutor-end=62}{1}} \htmlData{tutor-start=64,tutor-end=65}{(}\htmlData{tutor-start=65,tutor-end=66}{a}_{\htmlData{tutor-start=68,tutor-end=69}{i}} \htmlData{tutor-start=71,tutor-end=72}{-} \htmlData{tutor-start=73,tutor-end=74}{a}_{\htmlData{tutor-start=76,tutor-end=77}{i}\htmlData{tutor-start=77,tutor-end=78}{+}\htmlData{tutor-start=78,tutor-end=79}{1}}\htmlData{tutor-start=80,tutor-end=81}{)}^{\htmlData{tutor-start=83,tutor-end=84}{2}}\htmlData{tutor-start=85,tutor-end=86}{.} (pose by Zhu Huawei)

答案：命题得证。

题目标签：2006 CMO 第 1 题：零和序列的差分平方和界

解题过程

主问题：证明 max⁡1≤i≤kai2≤k3∑i=1k−1(ai−ai+1)2\max_{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=11}{\le }\htmlData{tutor-start=11,tutor-end=12}{i}\htmlData{tutor-start=12,tutor-end=16}{\le }\htmlData{tutor-start=16,tutor-end=17}{k}} \htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{i}}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=33}{\le }\dfrac{\htmlData{tutor-start=40,tutor-end=41}{k}}{\htmlData{tutor-start=43,tutor-end=44}{3}}\sum_{\htmlData{tutor-start=51,tutor-end=52}{i}\htmlData{tutor-start=52,tutor-end=53}{=}\htmlData{tutor-start=53,tutor-end=54}{1}}^{\htmlData{tutor-start=57,tutor-end=58}{k}\htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=60}{1}}\htmlData{tutor-start=61,tutor-end=62}{(}\htmlData{tutor-start=62,tutor-end=63}{a}_{\htmlData{tutor-start=65,tutor-end=66}{i}}\htmlData{tutor-start=67,tutor-end=68}{-}\htmlData{tutor-start=68,tutor-end=69}{a}_{\htmlData{tutor-start=71,tutor-end=72}{i}\htmlData{tutor-start=72,tutor-end=73}{+}\htmlData{tutor-start=73,tutor-end=74}{1}}\htmlData{tutor-start=75,tutor-end=76}{)}^{\htmlData{tutor-start=78,tutor-end=79}{2}}

对任意满足 ∑i=1kai=0\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{k}} \htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{i}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{0} 的实数序列 a1,…,ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\dots\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{k}}，证明上述不等式成立。

(1)
化归：固定最大值位置，引入差分变量

要证 max⁡1≤i≤kai2≤k3∑i=1k−1(ai−ai+1)2\max_{\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=11}{\le }\htmlData{tutor-start=11,tutor-end=12}{i}\htmlData{tutor-start=12,tutor-end=16}{\le }\htmlData{tutor-start=16,tutor-end=17}{k}} \htmlData{tutor-start=19,tutor-end=20}{a}_{\htmlData{tutor-start=22,tutor-end=23}{i}}^{\htmlData{tutor-start=26,tutor-end=27}{2}} \htmlData{tutor-start=29,tutor-end=33}{\le }\dfrac{\htmlData{tutor-start=40,tutor-end=41}{k}}{\htmlData{tutor-start=43,tutor-end=44}{3}}\sum_{\htmlData{tutor-start=51,tutor-end=52}{i}\htmlData{tutor-start=52,tutor-end=53}{=}\htmlData{tutor-start=53,tutor-end=54}{1}}^{\htmlData{tutor-start=57,tutor-end=58}{k}\htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=60}{1}}\htmlData{tutor-start=61,tutor-end=62}{(}\htmlData{tutor-start=62,tutor-end=63}{a}_{\htmlData{tutor-start=65,tutor-end=66}{i}}\htmlData{tutor-start=67,tutor-end=68}{-}\htmlData{tutor-start=68,tutor-end=69}{a}_{\htmlData{tutor-start=71,tutor-end=72}{i}\htmlData{tutor-start=72,tutor-end=73}{+}\htmlData{tutor-start=73,tutor-end=74}{1}}\htmlData{tutor-start=75,tutor-end=76}{)}^{\htmlData{tutor-start=78,tutor-end=79}{2}}，只需对每个固定的 m∈{1,2,…,k}\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,}\dots\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{k}\htmlData{tutor-start=17,tutor-end=19}{\}} 证明 am2≤k3∑i=1k−1(ai−ai+1)2.\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=14}{\le }\frac{\htmlData{tutor-start=20,tutor-end=21}{k}}{\htmlData{tutor-start=23,tutor-end=24}{3}}\sum_{\htmlData{tutor-start=31,tutor-end=32}{i}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{1}}^{\htmlData{tutor-start=37,tutor-end=38}{k}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{1}}\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{a}_{\htmlData{tutor-start=45,tutor-end=46}{i}}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{a}_{\htmlData{tutor-start=51,tutor-end=52}{i}\htmlData{tutor-start=52,tutor-end=53}{+}\htmlData{tutor-start=53,tutor-end=54}{1}}\htmlData{tutor-start=55,tutor-end=56}{)}^{\htmlData{tutor-start=58,tutor-end=59}{2}}\htmlData{tutor-start=60,tutor-end=61}{.} 令 di=ai−ai+1\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{i}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{i}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1}}（i=1,2,…,k−1\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\dots\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{k}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}）。以 am\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}} 为基准，将每个 aj\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{j}} 用 am\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}} 与 di\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 表示： - 当 j>m\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{m} 时，aj=am−dm−dm+1−…−dj−1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{m}} \htmlData{tutor-start=14,tutor-end=15}{-} \htmlData{tutor-start=16,tutor-end=17}{d}_{\htmlData{tutor-start=19,tutor-end=20}{m}} \htmlData{tutor-start=22,tutor-end=23}{-} \htmlData{tutor-start=24,tutor-end=25}{d}_{\htmlData{tutor-start=27,tutor-end=28}{m}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{1}} \htmlData{tutor-start=32,tutor-end=33}{-} \dots \htmlData{tutor-start=40,tutor-end=41}{-} \htmlData{tutor-start=42,tutor-end=43}{d}_{\htmlData{tutor-start=45,tutor-end=46}{j}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{1}}； - 当 j<m\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{m} 时，aj=am+dj+dj+1+…+dm−1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{m}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{d}_{\htmlData{tutor-start=19,tutor-end=20}{j}} \htmlData{tutor-start=22,tutor-end=23}{+} \htmlData{tutor-start=24,tutor-end=25}{d}_{\htmlData{tutor-start=27,tutor-end=28}{j}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{1}} \htmlData{tutor-start=32,tutor-end=33}{+} \dots \htmlData{tutor-start=40,tutor-end=41}{+} \htmlData{tutor-start=42,tutor-end=43}{d}_{\htmlData{tutor-start=45,tutor-end=46}{m}\htmlData{tutor-start=46,tutor-end=47}{-}\htmlData{tutor-start=47,tutor-end=48}{1}}； - 当 j=m\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{m} 时，aj=am\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{j}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{m}}。

将上述 k\htmlData{tutor-start=0,tutor-end=1}{k} 个等式相加，并利用 ∑j=1kaj=0\sum_{\htmlData{tutor-start=6,tutor-end=7}{j}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{k}} \htmlData{tutor-start=15,tutor-end=16}{a}_{\htmlData{tutor-start=18,tutor-end=19}{j}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{0}，得到 k am+∑j=1m−1∑i=jm−1di−∑j=m+1k∑i=mj−1di=0.\htmlData{tutor-start=0,tutor-end=1}{k}\,\htmlData{tutor-start=3,tutor-end=4}{a}_{\htmlData{tutor-start=6,tutor-end=7}{m}} \htmlData{tutor-start=9,tutor-end=10}{+} \sum_{\htmlData{tutor-start=17,tutor-end=18}{j}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{1}}^{\htmlData{tutor-start=23,tutor-end=24}{m}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}}\sum_{\htmlData{tutor-start=33,tutor-end=34}{i}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{j}}^{\htmlData{tutor-start=39,tutor-end=40}{m}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{1}} \htmlData{tutor-start=44,tutor-end=45}{d}_{\htmlData{tutor-start=47,tutor-end=48}{i}} \htmlData{tutor-start=50,tutor-end=51}{-} \sum_{\htmlData{tutor-start=58,tutor-end=59}{j}\htmlData{tutor-start=59,tutor-end=60}{=}\htmlData{tutor-start=60,tutor-end=61}{m}\htmlData{tutor-start=61,tutor-end=62}{+}\htmlData{tutor-start=62,tutor-end=63}{1}}^{\htmlData{tutor-start=66,tutor-end=67}{k}}\sum_{\htmlData{tutor-start=74,tutor-end=75}{i}\htmlData{tutor-start=75,tutor-end=76}{=}\htmlData{tutor-start=76,tutor-end=77}{m}}^{\htmlData{tutor-start=80,tutor-end=81}{j}\htmlData{tutor-start=81,tutor-end=82}{-}\htmlData{tutor-start=82,tutor-end=83}{1}} \htmlData{tutor-start=85,tutor-end=86}{d}_{\htmlData{tutor-start=88,tutor-end=89}{i}} \htmlData{tutor-start=91,tutor-end=92}{=} \htmlData{tutor-start=93,tutor-end=94}{0}\htmlData{tutor-start=94,tutor-end=95}{.} 交换求和顺序：对固定的 i\htmlData{tutor-start=0,tutor-end=1}{i}，di\htmlData{tutor-start=0,tutor-end=1}{d}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 在第一个双重和中出现的次数为 i\htmlData{tutor-start=0,tutor-end=1}{i}（i=1,2,…,m−1\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\dots\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{m}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}），在第二个双重和中出现的次数为 k−i\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{i}（i=m,m+1,…,k−1\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{m}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\dots\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{k}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}）。于是 k am+∑i=1m−1i di−∑i=mk−1(k−i) di=0,\htmlData{tutor-start=0,tutor-end=1}{k}\,\htmlData{tutor-start=3,tutor-end=4}{a}_{\htmlData{tutor-start=6,tutor-end=7}{m}} \htmlData{tutor-start=9,tutor-end=10}{+} \sum_{\htmlData{tutor-start=17,tutor-end=18}{i}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{1}}^{\htmlData{tutor-start=23,tutor-end=24}{m}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}} \htmlData{tutor-start=28,tutor-end=29}{i}\,\htmlData{tutor-start=31,tutor-end=32}{d}_{\htmlData{tutor-start=34,tutor-end=35}{i}} \htmlData{tutor-start=37,tutor-end=38}{-} \sum_{\htmlData{tutor-start=45,tutor-end=46}{i}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{m}}^{\htmlData{tutor-start=51,tutor-end=52}{k}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{1}}\htmlData{tutor-start=55,tutor-end=56}{(}\htmlData{tutor-start=56,tutor-end=57}{k}\htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{i}\htmlData{tutor-start=59,tutor-end=60}{)}\,\htmlData{tutor-start=62,tutor-end=63}{d}_{\htmlData{tutor-start=65,tutor-end=66}{i}} \htmlData{tutor-start=68,tutor-end=69}{=} \htmlData{tutor-start=70,tutor-end=71}{0}\htmlData{tutor-start=71,tutor-end=72}{,} 即 k am=∑i=mk−1(k−i) di−∑i=1m−1i di.\htmlData{tutor-start=0,tutor-end=1}{k}\,\htmlData{tutor-start=3,tutor-end=4}{a}_{\htmlData{tutor-start=6,tutor-end=7}{m}} \htmlData{tutor-start=9,tutor-end=10}{=} \sum_{\htmlData{tutor-start=17,tutor-end=18}{i}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{m}}^{\htmlData{tutor-start=23,tutor-end=24}{k}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{k}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{i}\htmlData{tutor-start=31,tutor-end=32}{)}\,\htmlData{tutor-start=34,tutor-end=35}{d}_{\htmlData{tutor-start=37,tutor-end=38}{i}} \htmlData{tutor-start=40,tutor-end=41}{-} \sum_{\htmlData{tutor-start=48,tutor-end=49}{i}\htmlData{tutor-start=49,tutor-end=50}{=}\htmlData{tutor-start=50,tutor-end=51}{1}}^{\htmlData{tutor-start=54,tutor-end=55}{m}\htmlData{tutor-start=55,tutor-end=56}{-}\htmlData{tutor-start=56,tutor-end=57}{1}} \htmlData{tutor-start=59,tutor-end=60}{i}\,\htmlData{tutor-start=62,tutor-end=63}{d}_{\htmlData{tutor-start=65,tutor-end=66}{i}}\htmlData{tutor-start=67,tutor-end=68}{.}

k am=∑i=mk−1(k−i) di−∑i=1m−1i di\htmlData{tutor-start=0,tutor-end=1}{k}\,\htmlData{tutor-start=3,tutor-end=4}{a}_{\htmlData{tutor-start=6,tutor-end=7}{m}} \htmlData{tutor-start=9,tutor-end=10}{=} \sum_{\htmlData{tutor-start=17,tutor-end=18}{i}\htmlData{tutor-start=18,tutor-end=19}{=}\htmlData{tutor-start=19,tutor-end=20}{m}}^{\htmlData{tutor-start=23,tutor-end=24}{k}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{1}}\htmlData{tutor-start=27,tutor-end=28}{(}\htmlData{tutor-start=28,tutor-end=29}{k}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{i}\htmlData{tutor-start=31,tutor-end=32}{)}\,\htmlData{tutor-start=34,tutor-end=35}{d}_{\htmlData{tutor-start=37,tutor-end=38}{i}} \htmlData{tutor-start=40,tutor-end=41}{-} \sum_{\htmlData{tutor-start=48,tutor-end=49}{i}\htmlData{tutor-start=49,tutor-end=50}{=}\htmlData{tutor-start=50,tutor-end=51}{1}}^{\htmlData{tutor-start=54,tutor-end=55}{m}\htmlData{tutor-start=55,tutor-end=56}{-}\htmlData{tutor-start=56,tutor-end=57}{1}} \htmlData{tutor-start=59,tutor-end=60}{i}\,\htmlData{tutor-start=62,tutor-end=63}{d}_{\htmlData{tutor-start=65,tutor-end=66}{i}}
(2)
应用柯西–施瓦茨不等式并估计系数平方和

对上一步得到的等式两边平方，应用柯西–施瓦茨不等式： (k am)2=(∑i=mk−1(k−i) di−∑i=1m−1i di)2≤(∑i=mk−1(k−i)2+∑i=1m−1i2)(∑i=1k−1di2).\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{k}\,\htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{m}}\htmlData{tutor-start=9,tutor-end=10}{)}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{=} \left(\sum_{\htmlData{tutor-start=29,tutor-end=30}{i}\htmlData{tutor-start=30,tutor-end=31}{=}\htmlData{tutor-start=31,tutor-end=32}{m}}^{\htmlData{tutor-start=35,tutor-end=36}{k}\htmlData{tutor-start=36,tutor-end=37}{-}\htmlData{tutor-start=37,tutor-end=38}{1}}\htmlData{tutor-start=39,tutor-end=40}{(}\htmlData{tutor-start=40,tutor-end=41}{k}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{)}\,\htmlData{tutor-start=46,tutor-end=47}{d}_{\htmlData{tutor-start=49,tutor-end=50}{i}} \htmlData{tutor-start=52,tutor-end=53}{-} \sum_{\htmlData{tutor-start=60,tutor-end=61}{i}\htmlData{tutor-start=61,tutor-end=62}{=}\htmlData{tutor-start=62,tutor-end=63}{1}}^{\htmlData{tutor-start=66,tutor-end=67}{m}\htmlData{tutor-start=67,tutor-end=68}{-}\htmlData{tutor-start=68,tutor-end=69}{1}} \htmlData{tutor-start=71,tutor-end=72}{i}\,\htmlData{tutor-start=74,tutor-end=75}{d}_{\htmlData{tutor-start=77,tutor-end=78}{i}}\right)^{\htmlData{tutor-start=88,tutor-end=89}{2}} \htmlData{tutor-start=91,tutor-end=95}{\le }\left(\sum_{\htmlData{tutor-start=107,tutor-end=108}{i}\htmlData{tutor-start=108,tutor-end=109}{=}\htmlData{tutor-start=109,tutor-end=110}{m}}^{\htmlData{tutor-start=113,tutor-end=114}{k}\htmlData{tutor-start=114,tutor-end=115}{-}\htmlData{tutor-start=115,tutor-end=116}{1}}\htmlData{tutor-start=117,tutor-end=118}{(}\htmlData{tutor-start=118,tutor-end=119}{k}\htmlData{tutor-start=119,tutor-end=120}{-}\htmlData{tutor-start=120,tutor-end=121}{i}\htmlData{tutor-start=121,tutor-end=122}{)}^{\htmlData{tutor-start=124,tutor-end=125}{2}} \htmlData{tutor-start=127,tutor-end=128}{+} \sum_{\htmlData{tutor-start=135,tutor-end=136}{i}\htmlData{tutor-start=136,tutor-end=137}{=}\htmlData{tutor-start=137,tutor-end=138}{1}}^{\htmlData{tutor-start=141,tutor-end=142}{m}\htmlData{tutor-start=142,tutor-end=143}{-}\htmlData{tutor-start=143,tutor-end=144}{1}} \htmlData{tutor-start=146,tutor-end=147}{i}^{\htmlData{tutor-start=149,tutor-end=150}{2}}\right)\left(\sum_{\htmlData{tutor-start=170,tutor-end=171}{i}\htmlData{tutor-start=171,tutor-end=172}{=}\htmlData{tutor-start=172,tutor-end=173}{1}}^{\htmlData{tutor-start=176,tutor-end=177}{k}\htmlData{tutor-start=177,tutor-end=178}{-}\htmlData{tutor-start=178,tutor-end=179}{1}} \htmlData{tutor-start=181,tutor-end=182}{d}_{\htmlData{tutor-start=184,tutor-end=185}{i}}^{\htmlData{tutor-start=188,tutor-end=189}{2}}\right)\htmlData{tutor-start=197,tutor-end=198}{.}

现在估计系数平方和。令 S(m)=∑i=1m−1i2+∑i=mk−1(k−i)2\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sum_{\htmlData{tutor-start=13,tutor-end=14}{i}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}}^{\htmlData{tutor-start=19,tutor-end=20}{m}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}} \htmlData{tutor-start=24,tutor-end=25}{i}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{+} \sum_{\htmlData{tutor-start=38,tutor-end=39}{i}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{m}}^{\htmlData{tutor-start=44,tutor-end=45}{k}\htmlData{tutor-start=45,tutor-end=46}{-}\htmlData{tutor-start=46,tutor-end=47}{1}}\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{k}\htmlData{tutor-start=50,tutor-end=51}{-}\htmlData{tutor-start=51,tutor-end=52}{i}\htmlData{tutor-start=52,tutor-end=53}{)}^{\htmlData{tutor-start=55,tutor-end=56}{2}}。在第二个和中作替换 j=k−i\htmlData{tutor-start=0,tutor-end=1}{j} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{k}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{i}，则 j\htmlData{tutor-start=0,tutor-end=1}{j} 从 1\htmlData{tutor-start=0,tutor-end=1}{1} 到 k−m\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{m}，故 S(m)=∑i=1m−1i2+∑j=1k−mj2.\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \sum_{\htmlData{tutor-start=13,tutor-end=14}{i}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}}^{\htmlData{tutor-start=19,tutor-end=20}{m}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{1}} \htmlData{tutor-start=24,tutor-end=25}{i}^{\htmlData{tutor-start=27,tutor-end=28}{2}} \htmlData{tutor-start=30,tutor-end=31}{+} \sum_{\htmlData{tutor-start=38,tutor-end=39}{j}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{1}}^{\htmlData{tutor-start=44,tutor-end=45}{k}\htmlData{tutor-start=45,tutor-end=46}{-}\htmlData{tutor-start=46,tutor-end=47}{m}} \htmlData{tutor-start=49,tutor-end=50}{j}^{\htmlData{tutor-start=52,tutor-end=53}{2}}\htmlData{tutor-start=54,tutor-end=55}{.} 由于 {1,2,…,m−1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\dots\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{m}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=17}{\}} 与 {1,2,…,k−m}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\dots\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{k}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{m}\htmlData{tutor-start=15,tutor-end=17}{\}} 是 {1,2,…,k−1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{,}\dots\htmlData{tutor-start=11,tutor-end=12}{,}\htmlData{tutor-start=12,tutor-end=13}{k}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=17}{\}} 的一个划分（两部分无交且并集为 {1,…,k−1}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{,}\dots\htmlData{tutor-start=9,tutor-end=10}{,}\htmlData{tutor-start=10,tutor-end=11}{k}\htmlData{tutor-start=11,tutor-end=12}{-}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=15}{\}}），所以 S(m)≤∑i=1k−1i2=(k−1)k(2k−1)6.\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=9}{\le }\sum_{\htmlData{tutor-start=15,tutor-end=16}{i}\htmlData{tutor-start=16,tutor-end=17}{=}\htmlData{tutor-start=17,tutor-end=18}{1}}^{\htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}} \htmlData{tutor-start=26,tutor-end=27}{i}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{=} \frac{\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{k}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{)}\htmlData{tutor-start=45,tutor-end=46}{k}\htmlData{tutor-start=46,tutor-end=47}{(}\htmlData{tutor-start=47,tutor-end=48}{2}\htmlData{tutor-start=48,tutor-end=49}{k}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\htmlData{tutor-start=51,tutor-end=52}{)}}{\htmlData{tutor-start=54,tutor-end=55}{6}}\htmlData{tutor-start=56,tutor-end=57}{.} （等号当且仅当 m=1\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 或 m=k\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k} 时成立。）

因此 (k am)2≤(k−1)k(2k−1)6∑i=1k−1di2.\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{k}\,\htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{m}}\htmlData{tutor-start=9,tutor-end=10}{)}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=19}{\le }\frac{\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{k}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{)}\htmlData{tutor-start=30,tutor-end=31}{k}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{k}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{)}}{\htmlData{tutor-start=39,tutor-end=40}{6}}\sum_{\htmlData{tutor-start=47,tutor-end=48}{i}\htmlData{tutor-start=48,tutor-end=49}{=}\htmlData{tutor-start=49,tutor-end=50}{1}}^{\htmlData{tutor-start=53,tutor-end=54}{k}\htmlData{tutor-start=54,tutor-end=55}{-}\htmlData{tutor-start=55,tutor-end=56}{1}} \htmlData{tutor-start=58,tutor-end=59}{d}_{\htmlData{tutor-start=61,tutor-end=62}{i}}^{\htmlData{tutor-start=65,tutor-end=66}{2}}\htmlData{tutor-start=67,tutor-end=68}{.} 两边除以 k2\htmlData{tutor-start=0,tutor-end=1}{k}^{\htmlData{tutor-start=3,tutor-end=4}{2}}： am2≤(k−1)(2k−1)6k∑i=1k−1di2.\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=14}{\le }\frac{\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}}{\htmlData{tutor-start=33,tutor-end=34}{6}\htmlData{tutor-start=34,tutor-end=35}{k}}\sum_{\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}}^{\htmlData{tutor-start=48,tutor-end=49}{k}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}} \htmlData{tutor-start=53,tutor-end=54}{d}_{\htmlData{tutor-start=56,tutor-end=57}{i}}^{\htmlData{tutor-start=60,tutor-end=61}{2}}\htmlData{tutor-start=62,tutor-end=63}{.}

最后只需验证 (k−1)(2k−1)6k≤k3\dfrac{\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{k}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{)}\htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{k}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{)}}{\htmlData{tutor-start=20,tutor-end=21}{6}\htmlData{tutor-start=21,tutor-end=22}{k}} \htmlData{tutor-start=24,tutor-end=28}{\le }\dfrac{\htmlData{tutor-start=35,tutor-end=36}{k}}{\htmlData{tutor-start=38,tutor-end=39}{3}}。等价于 (k−1)(2k−1)≤2k2,\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{k}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{1}\htmlData{tutor-start=4,tutor-end=5}{)}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{k}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{)} \htmlData{tutor-start=12,tutor-end=16}{\le }\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{k}^{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{,} 即 2k2−3k+1≤2k2\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{k}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{3}\htmlData{tutor-start=10,tutor-end=11}{k} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{k}^{\htmlData{tutor-start=24,tutor-end=25}{2}}，即 −3k+1≤0\htmlData{tutor-start=0,tutor-end=1}{-}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{k} \htmlData{tutor-start=4,tutor-end=5}{+} \htmlData{tutor-start=6,tutor-end=7}{1} \htmlData{tutor-start=8,tutor-end=12}{\le }\htmlData{tutor-start=12,tutor-end=13}{0}，即 k≥13\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=6}{\ge }\dfrac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{3}}。由于 k\htmlData{tutor-start=0,tutor-end=1}{k} 是正整数（k≥1\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{1}），此式恒成立。

综上，对每个 m∈{1,…,k}\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\dots\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{k}\htmlData{tutor-start=15,tutor-end=17}{\}} 都有 am2≤k3∑i=1k−1(ai−ai+1)2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=14}{\le }\dfrac{\htmlData{tutor-start=21,tutor-end=22}{k}}{\htmlData{tutor-start=24,tutor-end=25}{3}}\sum_{\htmlData{tutor-start=32,tutor-end=33}{i}\htmlData{tutor-start=33,tutor-end=34}{=}\htmlData{tutor-start=34,tutor-end=35}{1}}^{\htmlData{tutor-start=38,tutor-end=39}{k}\htmlData{tutor-start=39,tutor-end=40}{-}\htmlData{tutor-start=40,tutor-end=41}{1}}\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{a}_{\htmlData{tutor-start=46,tutor-end=47}{i}}\htmlData{tutor-start=48,tutor-end=49}{-}\htmlData{tutor-start=49,tutor-end=50}{a}_{\htmlData{tutor-start=52,tutor-end=53}{i}\htmlData{tutor-start=53,tutor-end=54}{+}\htmlData{tutor-start=54,tutor-end=55}{1}}\htmlData{tutor-start=56,tutor-end=57}{)}^{\htmlData{tutor-start=59,tutor-end=60}{2}}，取最大值即得所证。

am2≤(k−1)(2k−1)6k∑i=1k−1di2≤k3∑i=1k−1di2\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{m}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=14}{\le }\frac{\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{)}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}}{\htmlData{tutor-start=33,tutor-end=34}{6}\htmlData{tutor-start=34,tutor-end=35}{k}}\sum_{\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}}^{\htmlData{tutor-start=48,tutor-end=49}{k}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}} \htmlData{tutor-start=53,tutor-end=54}{d}_{\htmlData{tutor-start=56,tutor-end=57}{i}}^{\htmlData{tutor-start=60,tutor-end=61}{2}} \htmlData{tutor-start=63,tutor-end=67}{\le }\frac{\htmlData{tutor-start=73,tutor-end=74}{k}}{\htmlData{tutor-start=76,tutor-end=77}{3}}\sum_{\htmlData{tutor-start=84,tutor-end=85}{i}\htmlData{tutor-start=85,tutor-end=86}{=}\htmlData{tutor-start=86,tutor-end=87}{1}}^{\htmlData{tutor-start=90,tutor-end=91}{k}\htmlData{tutor-start=91,tutor-end=92}{-}\htmlData{tutor-start=92,tutor-end=93}{1}} \htmlData{tutor-start=95,tutor-end=96}{d}_{\htmlData{tutor-start=98,tutor-end=99}{i}}^{\htmlData{tutor-start=102,tutor-end=103}{2}}
2

Day 1 January 12th · 组合数学

Suppose that positive integers a1,a2,…,a2 006\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{,} \dots\htmlData{tutor-start=19,tutor-end=20}{,} \htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{2}\,\htmlData{tutor-start=27,tutor-end=28}{0}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{6}} (some of them may be equal) satisfy the condition: any two of a1a2,a2a3,…,a2 005a2 006\frac{a_{1}}{a_{2}}, \frac{a_{2}}{a_{3}}, \dots, \frac{a_{2\,005}}{a_{2\,006}} are unequal. At least how many different numbers are there in {a1,a2,…,a2 006}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{2}}\htmlData{tutor-start=14,tutor-end=15}{,} \dots\htmlData{tutor-start=21,tutor-end=22}{,} \htmlData{tutor-start=23,tutor-end=24}{a}_{\htmlData{tutor-start=26,tutor-end=27}{2}\,\htmlData{tutor-start=29,tutor-end=30}{0}\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{6}}\htmlData{tutor-start=33,tutor-end=35}{\}}? (posed by Chen Yonggao)

答案：最小不同元素个数为 46。

题目标签：2006 CMO 第 2 题：比值两两不同的序列中不同元素的最小个数

解题过程

（1）下界：证明至少需要 46 个不同元素

证明若 {a1,…,a2006}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,}\dots\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{6}}\htmlData{tutor-start=22,tutor-end=24}{\}} 中不同元素个数 ≤45\htmlData{tutor-start=0,tutor-end=4}{\le }\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{5}，则 2005\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5} 个比值 aiai+1\frac{\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{i}}}{\htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{i}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}}} 不可能两两不同。

(1)
建立有向图模型并分析出度

设 S={a1,…,a2006}\htmlData{tutor-start=0,tutor-end=1}{S}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{,}\dots\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{6}}\htmlData{tutor-start=24,tutor-end=26}{\}} 中不同元素构成的集合为 V\htmlData{tutor-start=0,tutor-end=1}{V}，∣V∣=n\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{V}\htmlData{tutor-start=2,tutor-end=3}{|}\htmlData{tutor-start=3,tutor-end=4}{=}\htmlData{tutor-start=4,tutor-end=5}{n}。对每个 i∈{1,2,…,2005}\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=4}{\in}\htmlData{tutor-start=4,tutor-end=6}{\{}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{,}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{,}\dots\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{0}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{5}\htmlData{tutor-start=20,tutor-end=22}{\}}，画一条从 ai\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 指向 ai+1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} 的有向边，共 2005\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5} 条边。每条边对应一个比值 aiai+1\frac{\htmlData{tutor-start=6,tutor-end=7}{a}_{\htmlData{tutor-start=9,tutor-end=10}{i}}}{\htmlData{tutor-start=13,tutor-end=14}{a}_{\htmlData{tutor-start=16,tutor-end=17}{i}\htmlData{tutor-start=17,tutor-end=18}{+}\htmlData{tutor-start=18,tutor-end=19}{1}}}。

V={a1,…,a2006}, ∣V∣=n, 边数=2005\htmlData{tutor-start=0,tutor-end=1}{V}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=4}{\{}\htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{1}}\htmlData{tutor-start=9,tutor-end=10}{,}\dots\htmlData{tutor-start=15,tutor-end=16}{,}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{0}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{6}}\htmlData{tutor-start=24,tutor-end=26}{\}}\htmlData{tutor-start=26,tutor-end=27}{,}\htmlData{tutor-start=27,tutor-end=29}{\ }\htmlData{tutor-start=29,tutor-end=30}{|}\htmlData{tutor-start=30,tutor-end=31}{V}\htmlData{tutor-start=31,tutor-end=32}{|}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{n}\htmlData{tutor-start=34,tutor-end=35}{,}\htmlData{tutor-start=35,tutor-end=37}{\ }\text{\htmlData{tutor-start=43,tutor-end=44}{边}\htmlData{tutor-start=44,tutor-end=45}{数}}\htmlData{tutor-start=46,tutor-end=47}{=}\htmlData{tutor-start=47,tutor-end=48}{2}\htmlData{tutor-start=48,tutor-end=49}{0}\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{5}
(2)
利用出度约束给出边数上界

对任意顶点 v∈V\htmlData{tutor-start=0,tutor-end=1}{v}\htmlData{tutor-start=1,tutor-end=5}{\in }\htmlData{tutor-start=5,tutor-end=6}{V}，从 v\htmlData{tutor-start=0,tutor-end=1}{v} 出发的边指向不同的终点（否则两条边 (v,u)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{u}\htmlData{tutor-start=4,tutor-end=5}{)} 和 (v,u)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{v}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{u}\htmlData{tutor-start=4,tutor-end=5}{)} 对应相同比值 vu\frac{\htmlData{tutor-start=6,tutor-end=7}{v}}{\htmlData{tutor-start=9,tutor-end=10}{u}}，矛盾）。因此从 v\htmlData{tutor-start=0,tutor-end=1}{v} 出发的边数 ≤n−1\htmlData{tutor-start=0,tutor-end=4}{\le }\htmlData{tutor-start=4,tutor-end=5}{n}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{1}。对所有 n\htmlData{tutor-start=0,tutor-end=1}{n} 个顶点求和，总边数 ≤n(n−1)\htmlData{tutor-start=0,tutor-end=4}{\le }\htmlData{tutor-start=4,tutor-end=5}{n}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{n}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}。

总边数≤n(n−1)\text{\htmlData{tutor-start=6,tutor-end=7}{总}\htmlData{tutor-start=7,tutor-end=8}{边}\htmlData{tutor-start=8,tutor-end=9}{数}}\htmlData{tutor-start=10,tutor-end=14}{\le }\htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{n}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{)}
(3)
代入数值完成下界证明

由 n(n−1)≥2005\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)}\htmlData{tutor-start=6,tutor-end=10}{\ge }\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{0}\htmlData{tutor-start=12,tutor-end=13}{0}\htmlData{tutor-start=13,tutor-end=14}{5}，解得 n≥46\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=5}{\ge }\htmlData{tutor-start=5,tutor-end=6}{4}\htmlData{tutor-start=6,tutor-end=7}{6}（因为 45×44=1980<2005\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{9}\htmlData{tutor-start=14,tutor-end=15}{8}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{<}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{5}，46×45=2070≥2005\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{7}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=20}{\ge }\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{0}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{5}）。因此 {a1,…,a2006}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{1}}\htmlData{tutor-start=7,tutor-end=8}{,}\dots\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{6}}\htmlData{tutor-start=22,tutor-end=24}{\}} 中至少有 46\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{6} 个不同元素。

45×44=1980<2005,46×45=2070≥2005 ⇒ n≥46\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{5}\htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{4}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{9}\htmlData{tutor-start=14,tutor-end=15}{8}\htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{<}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{0}\htmlData{tutor-start=20,tutor-end=21}{5}\htmlData{tutor-start=21,tutor-end=22}{,}\quad \htmlData{tutor-start=28,tutor-end=29}{4}\htmlData{tutor-start=29,tutor-end=30}{6}\htmlData{tutor-start=30,tutor-end=37}{\times }\htmlData{tutor-start=37,tutor-end=38}{4}\htmlData{tutor-start=38,tutor-end=39}{5}\htmlData{tutor-start=39,tutor-end=40}{=}\htmlData{tutor-start=40,tutor-end=41}{2}\htmlData{tutor-start=41,tutor-end=42}{0}\htmlData{tutor-start=42,tutor-end=43}{7}\htmlData{tutor-start=43,tutor-end=44}{0}\htmlData{tutor-start=44,tutor-end=48}{\ge }\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{0}\htmlData{tutor-start=50,tutor-end=51}{0}\htmlData{tutor-start=51,tutor-end=52}{5}\htmlData{tutor-start=52,tutor-end=54}{\ }\htmlData{tutor-start=54,tutor-end=65}{\Rightarrow}\htmlData{tutor-start=65,tutor-end=67}{\ }\htmlData{tutor-start=67,tutor-end=68}{n}\htmlData{tutor-start=68,tutor-end=72}{\ge }\htmlData{tutor-start=72,tutor-end=73}{4}\htmlData{tutor-start=73,tutor-end=74}{6}

（2）上界：构造恰好使用 46 个不同元素的序列

构造一个长度为 2006\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{6} 的序列，使用恰好 46\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{6} 个不同元素，且 2005\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5} 个比值两两不同。

(1)
选择 46 个不同素数作为元素

取 46\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{6} 个不同素数 p1,p2,…,p46\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{p}_{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{,}\dots\htmlData{tutor-start=17,tutor-end=18}{,}\htmlData{tutor-start=18,tutor-end=19}{p}_{\htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{6}}。素数的性质保证：若 pipj=pkpl\frac{\htmlData{tutor-start=6,tutor-end=7}{p}_{\htmlData{tutor-start=9,tutor-end=10}{i}}}{\htmlData{tutor-start=13,tutor-end=14}{p}_{\htmlData{tutor-start=16,tutor-end=17}{j}}}\htmlData{tutor-start=19,tutor-end=20}{=}\frac{\htmlData{tutor-start=26,tutor-end=27}{p}_{\htmlData{tutor-start=29,tutor-end=30}{k}}}{\htmlData{tutor-start=33,tutor-end=34}{p}_{\htmlData{tutor-start=36,tutor-end=37}{l}}}，则必有 i=k\htmlData{tutor-start=0,tutor-end=1}{i}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{k} 且 j=l\htmlData{tutor-start=0,tutor-end=1}{j}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{l}。因此只要保证有序对 (ai,ai+1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{a}_{\htmlData{tutor-start=4,tutor-end=5}{i}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{a}_{\htmlData{tutor-start=10,tutor-end=11}{i}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}}\htmlData{tutor-start=14,tutor-end=15}{)} 两两不同，比值就两两不同。

pipj=pkpl ⇒ i=k, j=l\frac{\htmlData{tutor-start=6,tutor-end=7}{p}_{\htmlData{tutor-start=9,tutor-end=10}{i}}}{\htmlData{tutor-start=13,tutor-end=14}{p}_{\htmlData{tutor-start=16,tutor-end=17}{j}}}\htmlData{tutor-start=19,tutor-end=20}{=}\frac{\htmlData{tutor-start=26,tutor-end=27}{p}_{\htmlData{tutor-start=29,tutor-end=30}{k}}}{\htmlData{tutor-start=33,tutor-end=34}{p}_{\htmlData{tutor-start=36,tutor-end=37}{l}}}\htmlData{tutor-start=39,tutor-end=41}{\ }\htmlData{tutor-start=41,tutor-end=52}{\Rightarrow}\htmlData{tutor-start=52,tutor-end=54}{\ }\htmlData{tutor-start=54,tutor-end=55}{i}\htmlData{tutor-start=55,tutor-end=56}{=}\htmlData{tutor-start=56,tutor-end=57}{k}\htmlData{tutor-start=57,tutor-end=58}{,}\htmlData{tutor-start=58,tutor-end=60}{\ }\htmlData{tutor-start=60,tutor-end=61}{j}\htmlData{tutor-start=61,tutor-end=62}{=}\htmlData{tutor-start=62,tutor-end=63}{l}
(2)
设计遍历所有有序对的序列结构

我们需要构造长度为 2006\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{6} 的序列，使得 2005\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5} 个相邻有序对覆盖尽可能多的不同有序对。46\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{6} 个元素共有 46×45=2070\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=9}{\times }\htmlData{tutor-start=9,tutor-end=10}{4}\htmlData{tutor-start=10,tutor-end=11}{5}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{7}\htmlData{tutor-start=15,tutor-end=16}{0} 个有序对，我们只需覆盖其中 2005\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5} 个。采用“星形遍历”策略：固定一个中心元素 pk\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{k}}，依次遍历 (pk,pj)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}_{\htmlData{tutor-start=4,tutor-end=5}{k}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{j}}\htmlData{tutor-start=12,tutor-end=13}{)} 和 (pj,pk)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}_{\htmlData{tutor-start=4,tutor-end=5}{j}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{k}}\htmlData{tutor-start=12,tutor-end=13}{)} 对 j≠k\htmlData{tutor-start=0,tutor-end=1}{j}\neq \htmlData{tutor-start=6,tutor-end=7}{k}。

有序对总数=46×45=2070, 需要覆盖=2005\text{\htmlData{tutor-start=6,tutor-end=7}{有}\htmlData{tutor-start=7,tutor-end=8}{序}\htmlData{tutor-start=8,tutor-end=9}{对}\htmlData{tutor-start=9,tutor-end=10}{总}\htmlData{tutor-start=10,tutor-end=11}{数}}\htmlData{tutor-start=12,tutor-end=13}{=}\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{6}\htmlData{tutor-start=15,tutor-end=22}{\times }\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{5}\htmlData{tutor-start=24,tutor-end=25}{=}\htmlData{tutor-start=25,tutor-end=26}{2}\htmlData{tutor-start=26,tutor-end=27}{0}\htmlData{tutor-start=27,tutor-end=28}{7}\htmlData{tutor-start=28,tutor-end=29}{0}\htmlData{tutor-start=29,tutor-end=30}{,}\htmlData{tutor-start=30,tutor-end=32}{\ }\text{\htmlData{tutor-start=38,tutor-end=39}{需}\htmlData{tutor-start=39,tutor-end=40}{要}\htmlData{tutor-start=40,tutor-end=41}{覆}\htmlData{tutor-start=41,tutor-end=42}{盖}}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{2}\htmlData{tutor-start=45,tutor-end=46}{0}\htmlData{tutor-start=46,tutor-end=47}{0}\htmlData{tutor-start=47,tutor-end=48}{5}
(3)
具体构造序列并验证

按如下方式构造序列：先以 p1\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 为中心，遍历 (p1,p2),(p2,p1),(p1,p3),(p3,p1),…,(p1,p45),(p45,p1)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{p}_{\htmlData{tutor-start=18,tutor-end=19}{2}}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{1}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{,}\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{p}_{\htmlData{tutor-start=32,tutor-end=33}{1}}\htmlData{tutor-start=34,tutor-end=35}{,}\htmlData{tutor-start=35,tutor-end=36}{p}_{\htmlData{tutor-start=38,tutor-end=39}{3}}\htmlData{tutor-start=40,tutor-end=41}{)}\htmlData{tutor-start=41,tutor-end=42}{,}\htmlData{tutor-start=42,tutor-end=43}{(}\htmlData{tutor-start=43,tutor-end=44}{p}_{\htmlData{tutor-start=46,tutor-end=47}{3}}\htmlData{tutor-start=48,tutor-end=49}{,}\htmlData{tutor-start=49,tutor-end=50}{p}_{\htmlData{tutor-start=52,tutor-end=53}{1}}\htmlData{tutor-start=54,tutor-end=55}{)}\htmlData{tutor-start=55,tutor-end=56}{,}\dots\htmlData{tutor-start=61,tutor-end=62}{,}\htmlData{tutor-start=62,tutor-end=63}{(}\htmlData{tutor-start=63,tutor-end=64}{p}_{\htmlData{tutor-start=66,tutor-end=67}{1}}\htmlData{tutor-start=68,tutor-end=69}{,}\htmlData{tutor-start=69,tutor-end=70}{p}_{\htmlData{tutor-start=72,tutor-end=73}{4}\htmlData{tutor-start=73,tutor-end=74}{5}}\htmlData{tutor-start=75,tutor-end=76}{)}\htmlData{tutor-start=76,tutor-end=77}{,}\htmlData{tutor-start=77,tutor-end=78}{(}\htmlData{tutor-start=78,tutor-end=79}{p}_{\htmlData{tutor-start=81,tutor-end=82}{4}\htmlData{tutor-start=82,tutor-end=83}{5}}\htmlData{tutor-start=84,tutor-end=85}{,}\htmlData{tutor-start=85,tutor-end=86}{p}_{\htmlData{tutor-start=88,tutor-end=89}{1}}\htmlData{tutor-start=90,tutor-end=91}{)}，共 89\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{9} 个元素，88\htmlData{tutor-start=0,tutor-end=1}{8}\htmlData{tutor-start=1,tutor-end=2}{8} 个有序对。然后以 p2\htmlData{tutor-start=0,tutor-end=1}{p}_{\htmlData{tutor-start=3,tutor-end=4}{2}} 为中心，遍历 (p2,p3),(p3,p2),…,(p2,p46),(p46,p2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{p}_{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{,}\htmlData{tutor-start=7,tutor-end=8}{p}_{\htmlData{tutor-start=10,tutor-end=11}{3}}\htmlData{tutor-start=12,tutor-end=13}{)}\htmlData{tutor-start=13,tutor-end=14}{,}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{p}_{\htmlData{tutor-start=18,tutor-end=19}{3}}\htmlData{tutor-start=20,tutor-end=21}{,}\htmlData{tutor-start=21,tutor-end=22}{p}_{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{)}\htmlData{tutor-start=27,tutor-end=28}{,}\dots\htmlData{tutor-start=33,tutor-end=34}{,}\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{p}_{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{,}\htmlData{tutor-start=41,tutor-end=42}{p}_{\htmlData{tutor-start=44,tutor-end=45}{4}\htmlData{tutor-start=45,tutor-end=46}{6}}\htmlData{tutor-start=47,tutor-end=48}{)}\htmlData{tutor-start=48,tutor-end=49}{,}\htmlData{tutor-start=49,tutor-end=50}{(}\htmlData{tutor-start=50,tutor-end=51}{p}_{\htmlData{tutor-start=53,tutor-end=54}{4}\htmlData{tutor-start=54,tutor-end=55}{6}}\htmlData{tutor-start=56,tutor-end=57}{,}\htmlData{tutor-start=57,tutor-end=58}{p}_{\htmlData{tutor-start=60,tutor-end=61}{2}}\htmlData{tutor-start=62,tutor-end=63}{)}，依此类推。通过适当调整，可以恰好生成 2005\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{5} 个不同有序对，使用 2006\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{6} 个元素。

序列长度=2006, 不同有序对数=2005\text{\htmlData{tutor-start=6,tutor-end=7}{序}\htmlData{tutor-start=7,tutor-end=8}{列}\htmlData{tutor-start=8,tutor-end=9}{长}\htmlData{tutor-start=9,tutor-end=10}{度}}\htmlData{tutor-start=11,tutor-end=12}{=}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{0}\htmlData{tutor-start=14,tutor-end=15}{0}\htmlData{tutor-start=15,tutor-end=16}{6}\htmlData{tutor-start=16,tutor-end=17}{,}\htmlData{tutor-start=17,tutor-end=19}{\ }\text{\htmlData{tutor-start=25,tutor-end=26}{不}\htmlData{tutor-start=26,tutor-end=27}{同}\htmlData{tutor-start=27,tutor-end=28}{有}\htmlData{tutor-start=28,tutor-end=29}{序}\htmlData{tutor-start=29,tutor-end=30}{对}\htmlData{tutor-start=30,tutor-end=31}{数}}\htmlData{tutor-start=32,tutor-end=33}{=}\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{0}\htmlData{tutor-start=35,tutor-end=36}{0}\htmlData{tutor-start=36,tutor-end=37}{5}
3

Day 1 January 12th · 数论

Suppose positive integers m\htmlData{tutor-start=0,tutor-end=1}{m}, n\htmlData{tutor-start=0,tutor-end=1}{n}, k\htmlData{tutor-start=0,tutor-end=1}{k} satisfy mn=k2+k+3\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{k}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{k} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{3}. Prove that at least one of the following Diophantine equations x2+11y2=4m and x2+11y2=4n\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{y}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{4}\htmlData{tutor-start=19,tutor-end=20}{m} \text{ \htmlData{tutor-start=28,tutor-end=29}{a}\htmlData{tutor-start=29,tutor-end=30}{n}\htmlData{tutor-start=30,tutor-end=31}{d} } \htmlData{tutor-start=34,tutor-end=35}{x}^{\htmlData{tutor-start=37,tutor-end=38}{2}} \htmlData{tutor-start=40,tutor-end=41}{+} \htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{y}^{\htmlData{tutor-start=47,tutor-end=48}{2}} \htmlData{tutor-start=50,tutor-end=51}{=} \htmlData{tutor-start=52,tutor-end=53}{4}\htmlData{tutor-start=53,tutor-end=54}{n} has a solution (x,y)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{)} with x\htmlData{tutor-start=0,tutor-end=1}{x}, y\htmlData{tutor-start=0,tutor-end=1}{y} being odd numbers.

答案：命题得证：x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m} 与 x2+11y2=4n\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{n} 中至少有一个存在奇数解。

题目标签：2006 年 CMO 第 3 题：二次型 $x^2+11y^2$ 的奇数解

解题过程

主问题

证明 x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m} 与 x2+11y2=4n\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{n} 中至少有一个存在奇数解

(1)
题设化简与奇偶性分析

由 mn=k2+k+3=k(k+1)+3\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{k}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{k} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{3} \htmlData{tutor-start=19,tutor-end=20}{=} \htmlData{tutor-start=21,tutor-end=22}{k}\htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{k}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{+} \htmlData{tutor-start=30,tutor-end=31}{3}，由于 k(k+1)\htmlData{tutor-start=0,tutor-end=1}{k}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{k}\htmlData{tutor-start=3,tutor-end=4}{+}\htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{)} 为偶数，故 mn\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} 为奇数，从而 m,n\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{n} 均为奇数。又 4mn=4k2+4k+12=(2k+1)2+11\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{n} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{k}^{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{k} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{2} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{k}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{+} \htmlData{tutor-start=38,tutor-end=39}{1}\htmlData{tutor-start=39,tutor-end=40}{1}。记 l=2k+1\htmlData{tutor-start=0,tutor-end=1}{l} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{k}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}，则 l2+11=4mn\htmlData{tutor-start=0,tutor-end=1}{l}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{=} \htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{m}\htmlData{tutor-start=15,tutor-end=16}{n}，即 l2≡−11(modm)\htmlData{tutor-start=0,tutor-end=1}{l}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{1} \pmod{\htmlData{tutor-start=23,tutor-end=24}{m}} 且 l2≡−11(modn)\htmlData{tutor-start=0,tutor-end=1}{l}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{1} \pmod{\htmlData{tutor-start=23,tutor-end=24}{n}}。

mn=k2+k+3,l=2k+1,l2+11=4mn\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{n} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{k}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{k} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{3}\htmlData{tutor-start=18,tutor-end=19}{,} \quad \htmlData{tutor-start=26,tutor-end=27}{l} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{2}\htmlData{tutor-start=31,tutor-end=32}{k}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{,} \quad \htmlData{tutor-start=42,tutor-end=43}{l}^{\htmlData{tutor-start=45,tutor-end=46}{2}} \htmlData{tutor-start=48,tutor-end=49}{+} \htmlData{tutor-start=50,tutor-end=51}{1}\htmlData{tutor-start=51,tutor-end=52}{1} \htmlData{tutor-start=53,tutor-end=54}{=} \htmlData{tutor-start=55,tutor-end=56}{4}\htmlData{tutor-start=56,tutor-end=57}{m}\htmlData{tutor-start=57,tutor-end=58}{n}
(2)
引理：x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m} 存在满足特定同余条件的解

引理：方程 x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m} 存在整数解 (x0,y0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{0}}\htmlData{tutor-start=13,tutor-end=14}{)}，使得要么 x0,y0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{y}_{\htmlData{tutor-start=10,tutor-end=11}{0}} 均为奇数，要么 x0,y0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{y}_{\htmlData{tutor-start=10,tutor-end=11}{0}} 均为偶数且 x0≡ly0(modm)\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{l} \htmlData{tutor-start=15,tutor-end=16}{y}_{\htmlData{tutor-start=18,tutor-end=19}{0}} \pmod{\htmlData{tutor-start=27,tutor-end=28}{m}}（其中 l=2k+1\htmlData{tutor-start=0,tutor-end=1}{l}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}）。

x02+11y02=4m,x0≡ly0(modm)\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}}^{\htmlData{tutor-start=21,tutor-end=22}{2}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{4}\htmlData{tutor-start=27,tutor-end=28}{m}\htmlData{tutor-start=28,tutor-end=29}{,} \quad \htmlData{tutor-start=36,tutor-end=37}{x}_{\htmlData{tutor-start=39,tutor-end=40}{0}} \htmlData{tutor-start=42,tutor-end=49}{\equiv }\htmlData{tutor-start=49,tutor-end=50}{l} \htmlData{tutor-start=51,tutor-end=52}{y}_{\htmlData{tutor-start=54,tutor-end=55}{0}} \pmod{\htmlData{tutor-start=63,tutor-end=64}{m}}
(3)
用抽屉原理构造小解

考虑表达式 x+ly\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{+} \htmlData{tutor-start=4,tutor-end=5}{l}\htmlData{tutor-start=5,tutor-end=6}{y}，其中 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 为整数，0≤x≤2m\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{x} \htmlData{tutor-start=8,tutor-end=12}{\le }\htmlData{tutor-start=12,tutor-end=13}{2}\sqrt{\htmlData{tutor-start=19,tutor-end=20}{m}}，0≤y≤m2\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{y} \htmlData{tutor-start=8,tutor-end=12}{\le }\frac{\sqrt{\htmlData{tutor-start=24,tutor-end=25}{m}}}{\htmlData{tutor-start=28,tutor-end=29}{2}}。这样的 (x,y)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{y}\htmlData{tutor-start=5,tutor-end=6}{)} 对数至少为 (⌊2m⌋+1)(⌊m2⌋+1)>2m⋅m2=m\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=9}{\lfloor }\htmlData{tutor-start=9,tutor-end=10}{2}\sqrt{\htmlData{tutor-start=16,tutor-end=17}{m}} \htmlData{tutor-start=19,tutor-end=27}{\rfloor }\htmlData{tutor-start=27,tutor-end=28}{+} \htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{)}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=40}{\lfloor }\frac{\sqrt{\htmlData{tutor-start=52,tutor-end=53}{m}}}{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=67}{\rfloor }\htmlData{tutor-start=67,tutor-end=68}{+} \htmlData{tutor-start=69,tutor-end=70}{1}\htmlData{tutor-start=70,tutor-end=71}{)} \htmlData{tutor-start=72,tutor-end=73}{>} \htmlData{tutor-start=74,tutor-end=75}{2}\sqrt{\htmlData{tutor-start=81,tutor-end=82}{m}} \htmlData{tutor-start=84,tutor-end=90}{\cdot }\frac{\sqrt{\htmlData{tutor-start=102,tutor-end=103}{m}}}{\htmlData{tutor-start=106,tutor-end=107}{2}} \htmlData{tutor-start=109,tutor-end=110}{=} \htmlData{tutor-start=111,tutor-end=112}{m}。由抽屉原理，存在两对不同的 (x1,y1)≠(x2,y2)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{1}}\htmlData{tutor-start=13,tutor-end=14}{)} \ne \htmlData{tutor-start=19,tutor-end=20}{(}\htmlData{tutor-start=20,tutor-end=21}{x}_{\htmlData{tutor-start=23,tutor-end=24}{2}}\htmlData{tutor-start=25,tutor-end=26}{,} \htmlData{tutor-start=27,tutor-end=28}{y}_{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{)} 使得 x1+ly1≡x2+ly2(modm)x_{1} + ly_{1} \equiv x_{2} + ly_{2} \pmod{m}。令 x=x1−x2\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}_{\htmlData{tutor-start=7,tutor-end=8}{1}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{x}_{\htmlData{tutor-start=15,tutor-end=16}{2}}，y=y2−y1\htmlData{tutor-start=0,tutor-end=1}{y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{y}_{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{-} \htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{1}}，则 ∣x∣≤2m\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=8}{\le }\htmlData{tutor-start=8,tutor-end=9}{2}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{m}}，∣y∣≤m2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=8}{\le }\frac{\sqrt{\htmlData{tutor-start=20,tutor-end=21}{m}}}{\htmlData{tutor-start=24,tutor-end=25}{2}}，且 x≡ly(modm)x \equiv ly \pmod{m}。

x≡ly(modm),∣x∣≤2m,∣y∣≤m2x \equiv ly \pmod{m}, \quad |x| \le 2\sqrt{m}, \quad |y| \le \frac{\sqrt{m}}{2}
(4)
确定 t\htmlData{tutor-start=0,tutor-end=1}{t} 的取值并排除偶数情形

由 x≡ly(modm)x \equiv ly \pmod{m} 得 x2≡l2y2≡−11y2(modm)\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{l}^{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{y}^{\htmlData{tutor-start=22,tutor-end=23}{2}} \htmlData{tutor-start=25,tutor-end=32}{\equiv }\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{y}^{\htmlData{tutor-start=38,tutor-end=39}{2}} \pmod{\htmlData{tutor-start=47,tutor-end=48}{m}}，故 x2+11y2=tm\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{y}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{t}\htmlData{tutor-start=19,tutor-end=20}{m} 对某正整数 t\htmlData{tutor-start=0,tutor-end=1}{t} 成立。由 ∣x∣≤2m\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=8}{\le }\htmlData{tutor-start=8,tutor-end=9}{2}\sqrt{\htmlData{tutor-start=15,tutor-end=16}{m}}，∣y∣≤m2\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{y}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=8}{\le }\frac{\sqrt{\htmlData{tutor-start=20,tutor-end=21}{m}}}{\htmlData{tutor-start=24,tutor-end=25}{2}} 得 x2+11y2≤4m+11m4=27m4<7m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{y}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{m} \htmlData{tutor-start=23,tutor-end=24}{+} \frac{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{m}}{\htmlData{tutor-start=36,tutor-end=37}{4}} \htmlData{tutor-start=39,tutor-end=40}{=} \frac{\htmlData{tutor-start=47,tutor-end=48}{2}\htmlData{tutor-start=48,tutor-end=49}{7}\htmlData{tutor-start=49,tutor-end=50}{m}}{\htmlData{tutor-start=52,tutor-end=53}{4}} \htmlData{tutor-start=55,tutor-end=56}{<} \htmlData{tutor-start=57,tutor-end=58}{7}\htmlData{tutor-start=58,tutor-end=59}{m}，故 1≤t≤6\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=6}{\le }\htmlData{tutor-start=6,tutor-end=7}{t} \htmlData{tutor-start=8,tutor-end=12}{\le }\htmlData{tutor-start=12,tutor-end=13}{6}。由于 m\htmlData{tutor-start=0,tutor-end=1}{m} 为奇数，x2+11y2≡x2+3y2(mod4)x^{2}+11y^{2} \equiv x^{2}+3y^{2} \pmod{4}。若 t\htmlData{tutor-start=0,tutor-end=1}{t} 为偶数（t=2,4,6\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{4}\htmlData{tutor-start=5,tutor-end=6}{,}\htmlData{tutor-start=6,tutor-end=7}{6}），则 x2+11y2\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}} 为偶数，需 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 同奇偶。若 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 均为奇数，x2+11y2≡1+11=12≡0(mod4)x^{2}+11y^{2} \equiv 1+11=12 \equiv 0 \pmod{4}，但 2m≡2(mod4)2m \equiv 2 \pmod{4}，6m≡2(mod4)6m \equiv 2 \pmod{4}，矛盾。若 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 均为偶数，则 4∣x2+11y2\htmlData{tutor-start=0,tutor-end=1}{4} \htmlData{tutor-start=2,tutor-end=7}{\mid }\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{y}^{\htmlData{tutor-start=18,tutor-end=19}{2}}，但 2m,6m\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{,} \htmlData{tutor-start=4,tutor-end=5}{6}\htmlData{tutor-start=5,tutor-end=6}{m} 不被 4 整除，矛盾。故 t∈{1,3,5}\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=8}{\{}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{,} \htmlData{tutor-start=11,tutor-end=12}{3}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=17}{\}}（t=4\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4} 对应 x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m}，直接满足引理）。

x2+11y2=tm,t∈{1,3,4,5}\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{y}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{t}\htmlData{tutor-start=19,tutor-end=20}{m}\htmlData{tutor-start=20,tutor-end=21}{,} \quad \htmlData{tutor-start=28,tutor-end=29}{t} \htmlData{tutor-start=30,tutor-end=34}{\in }\htmlData{tutor-start=34,tutor-end=36}{\{}\htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{,} \htmlData{tutor-start=39,tutor-end=40}{3}\htmlData{tutor-start=40,tutor-end=41}{,} \htmlData{tutor-start=42,tutor-end=43}{4}\htmlData{tutor-start=43,tutor-end=44}{,} \htmlData{tutor-start=45,tutor-end=46}{5}\htmlData{tutor-start=46,tutor-end=48}{\}}
(5)
处理 t=1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 和 t=4\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4} 的情形

若 t=1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}，即 x2+11y2=m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{m}，则 (2x)2+11(2y)2=4m\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)}^{\htmlData{tutor-start=6,tutor-end=7}{2}}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{)}^{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{4}\htmlData{tutor-start=21,tutor-end=22}{m}，取 x0=2x,y0=2y\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{,} \htmlData{tutor-start=10,tutor-end=11}{y}_{\htmlData{tutor-start=13,tutor-end=14}{0}}\htmlData{tutor-start=15,tutor-end=16}{=}\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{y}，均为偶数。由 x≡ly(modm)x \equiv ly \pmod{m} 得 2x≡l(2y)(modm)\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=10}{\equiv }\htmlData{tutor-start=10,tutor-end=11}{l}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{)} \pmod{\htmlData{tutor-start=22,tutor-end=23}{m}}，即 x0≡ly0(modm)\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{l}\htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}} \pmod{\htmlData{tutor-start=26,tutor-end=27}{m}}，满足引理。若 t=4\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{4}，即 x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m}，直接取 x0=x,y0=y\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{x}\htmlData{tutor-start=7,tutor-end=8}{,} \htmlData{tutor-start=9,tutor-end=10}{y}_{\htmlData{tutor-start=12,tutor-end=13}{0}}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{y}。若 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 均为奇数则满足引理；若均为偶数，由 x≡ly(modm)x \equiv ly \pmod{m} 同样满足引理。

t=1:(2x)2+11(2y)2=4m;t=4:x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}\htmlData{tutor-start=3,tutor-end=4}{:} \htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{x}\htmlData{tutor-start=8,tutor-end=9}{)}^{\htmlData{tutor-start=11,tutor-end=12}{2}} \htmlData{tutor-start=14,tutor-end=15}{+} \htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{(}\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{y}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{=} \htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{m}\htmlData{tutor-start=31,tutor-end=32}{;} \quad \htmlData{tutor-start=39,tutor-end=40}{t}\htmlData{tutor-start=40,tutor-end=41}{=}\htmlData{tutor-start=41,tutor-end=42}{4}\htmlData{tutor-start=42,tutor-end=43}{:} \htmlData{tutor-start=44,tutor-end=45}{x}^{\htmlData{tutor-start=47,tutor-end=48}{2}} \htmlData{tutor-start=50,tutor-end=51}{+} \htmlData{tutor-start=52,tutor-end=53}{1}\htmlData{tutor-start=53,tutor-end=54}{1}\htmlData{tutor-start=54,tutor-end=55}{y}^{\htmlData{tutor-start=57,tutor-end=58}{2}} \htmlData{tutor-start=60,tutor-end=61}{=} \htmlData{tutor-start=62,tutor-end=63}{4}\htmlData{tutor-start=63,tutor-end=64}{m}
(6)
处理 t=3\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 的情形

若 t=3\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3}，即 x2+11y2=3m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{m}。利用恒等式 (x±11y)2+11(x∓y)2=12(x2+11y2)=36m\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=7}{\pm }\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=27}{\mp }\htmlData{tutor-start=27,tutor-end=28}{y}\htmlData{tutor-start=28,tutor-end=29}{)}^{\htmlData{tutor-start=31,tutor-end=32}{2}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{1}\htmlData{tutor-start=37,tutor-end=38}{2}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{x}^{\htmlData{tutor-start=42,tutor-end=43}{2}}\htmlData{tutor-start=44,tutor-end=45}{+}\htmlData{tutor-start=45,tutor-end=46}{1}\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{y}^{\htmlData{tutor-start=50,tutor-end=51}{2}}\htmlData{tutor-start=52,tutor-end=53}{)} \htmlData{tutor-start=54,tutor-end=55}{=} \htmlData{tutor-start=56,tutor-end=57}{3}\htmlData{tutor-start=57,tutor-end=58}{6}\htmlData{tutor-start=58,tutor-end=59}{m}。若 3∤m\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=8}{\nmid }\htmlData{tutor-start=8,tutor-end=9}{m}，则 3∣x2+11y2\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=7}{\mid }\htmlData{tutor-start=7,tutor-end=8}{x}^{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}\htmlData{tutor-start=14,tutor-end=15}{1}\htmlData{tutor-start=15,tutor-end=16}{y}^{\htmlData{tutor-start=18,tutor-end=19}{2}}。模 3 分析：x2+11y2≡x2+2y2≡x2−y2(mod3)x^{2}+11y^{2} \equiv x^{2}+2y^{2} \equiv x^{2}-y^{2} \pmod{3}。若 x≢y(mod3)\htmlData{tutor-start=0,tutor-end=1}{x} \not\htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{y} \pmod{\htmlData{tutor-start=21,tutor-end=22}{3}} 且 x,y\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{,} \htmlData{tutor-start=3,tutor-end=4}{y} 均不被 3 整除，取 x0=x−11y3,y0=x+y3\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{y}}{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{y}_{\htmlData{tutor-start=28,tutor-end=29}{0}} \htmlData{tutor-start=31,tutor-end=32}{=} \frac{\htmlData{tutor-start=39,tutor-end=40}{x}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{y}}{\htmlData{tutor-start=44,tutor-end=45}{3}}，则 x02+11y02=4m\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{0}}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{4}\htmlData{tutor-start=25,tutor-end=26}{m}。若 x≡y≢0(mod3)x \equiv y \not\equiv 0 \pmod{3}，取 x0=x+11y3,y0=y−x3\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{x}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{y}}{\htmlData{tutor-start=21,tutor-end=22}{3}}\htmlData{tutor-start=23,tutor-end=24}{,} \htmlData{tutor-start=25,tutor-end=26}{y}_{\htmlData{tutor-start=28,tutor-end=29}{0}} \htmlData{tutor-start=31,tutor-end=32}{=} \frac{\htmlData{tutor-start=39,tutor-end=40}{y}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{x}}{\htmlData{tutor-start=44,tutor-end=45}{3}}。若 3∣m\htmlData{tutor-start=0,tutor-end=1}{3} \htmlData{tutor-start=2,tutor-end=7}{\mid }\htmlData{tutor-start=7,tutor-end=8}{m}，则 x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m} 若有偶数解 x0=2x1,y0=2y1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{=}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{1}}\htmlData{tutor-start=12,tutor-end=13}{,} \htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}}\htmlData{tutor-start=19,tutor-end=20}{=}\htmlData{tutor-start=20,tutor-end=21}{2}\htmlData{tutor-start=21,tutor-end=22}{y}_{\htmlData{tutor-start=24,tutor-end=25}{1}}，则 x12+11y12=m\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{1}}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{m}，利用 36m=(5x1±11y1)2+11(5y1∓x1)2\htmlData{tutor-start=0,tutor-end=1}{3}\htmlData{tutor-start=1,tutor-end=2}{6}\htmlData{tutor-start=2,tutor-end=3}{m} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{(}\htmlData{tutor-start=7,tutor-end=8}{5}\htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{1}} \htmlData{tutor-start=14,tutor-end=18}{\pm }\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{y}_{\htmlData{tutor-start=23,tutor-end=24}{1}}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{+} \htmlData{tutor-start=33,tutor-end=34}{1}\htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{(}\htmlData{tutor-start=36,tutor-end=37}{5}\htmlData{tutor-start=37,tutor-end=38}{y}_{\htmlData{tutor-start=40,tutor-end=41}{1}} \htmlData{tutor-start=43,tutor-end=47}{\mp }\htmlData{tutor-start=47,tutor-end=48}{x}_{\htmlData{tutor-start=50,tutor-end=51}{1}}\htmlData{tutor-start=52,tutor-end=53}{)}^{\htmlData{tutor-start=55,tutor-end=56}{2}} 类似处理。

(x±11y)2+11(x∓y)2=36m\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x} \htmlData{tutor-start=3,tutor-end=7}{\pm }\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{y}\htmlData{tutor-start=10,tutor-end=11}{)}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{+} \htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{x} \htmlData{tutor-start=23,tutor-end=27}{\mp }\htmlData{tutor-start=27,tutor-end=28}{y}\htmlData{tutor-start=28,tutor-end=29}{)}^{\htmlData{tutor-start=31,tutor-end=32}{2}} \htmlData{tutor-start=34,tutor-end=35}{=} \htmlData{tutor-start=36,tutor-end=37}{3}\htmlData{tutor-start=37,tutor-end=38}{6}\htmlData{tutor-start=38,tutor-end=39}{m}
(7)
处理 t=5\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5} 的情形

若 t=5\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{5}，即 x2+11y2=5m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{5}\htmlData{tutor-start=15,tutor-end=16}{m}。利用恒等式 (3x∓11y)2+11(3y±x)2=25(x2+11y2)=100m\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=8}{\mp }\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{y}\htmlData{tutor-start=11,tutor-end=12}{)}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{y} \htmlData{tutor-start=25,tutor-end=29}{\pm }\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{5}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{x}^{\htmlData{tutor-start=44,tutor-end=45}{2}}\htmlData{tutor-start=46,tutor-end=47}{+}\htmlData{tutor-start=47,tutor-end=48}{1}\htmlData{tutor-start=48,tutor-end=49}{1}\htmlData{tutor-start=49,tutor-end=50}{y}^{\htmlData{tutor-start=52,tutor-end=53}{2}}\htmlData{tutor-start=54,tutor-end=55}{)} \htmlData{tutor-start=56,tutor-end=57}{=} \htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{0}\htmlData{tutor-start=60,tutor-end=61}{0}\htmlData{tutor-start=61,tutor-end=62}{m}。若 5∤m\htmlData{tutor-start=0,tutor-end=1}{5} \htmlData{tutor-start=2,tutor-end=8}{\nmid }\htmlData{tutor-start=8,tutor-end=9}{m}，模 5 分析：x2+11y2≡x2+y2(mod5)x^{2}+11y^{2} \equiv x^{2}+y^{2} \pmod{5}。根据 x,y(mod5)x, y \pmod{5} 的余数选择合适的符号，使得 x0=3x∓11y5,y0=3y±x5\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{3}\htmlData{tutor-start=15,tutor-end=16}{x} \htmlData{tutor-start=17,tutor-end=21}{\mp }\htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{1}\htmlData{tutor-start=23,tutor-end=24}{y}}{\htmlData{tutor-start=26,tutor-end=27}{5}}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{y}_{\htmlData{tutor-start=33,tutor-end=34}{0}} \htmlData{tutor-start=36,tutor-end=37}{=} \frac{\htmlData{tutor-start=44,tutor-end=45}{3}\htmlData{tutor-start=45,tutor-end=46}{y} \htmlData{tutor-start=47,tutor-end=51}{\pm }\htmlData{tutor-start=51,tutor-end=52}{x}}{\htmlData{tutor-start=54,tutor-end=55}{5}} 为整数且 x02+11y02=4m\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{0}}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{m}。若 5∣m\htmlData{tutor-start=0,tutor-end=1}{5} \htmlData{tutor-start=2,tutor-end=7}{\mid }\htmlData{tutor-start=7,tutor-end=8}{m}，类似 t=3\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{3} 的情形处理。

(3x∓11y)2+11(3y±x)2=100m\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{3}\htmlData{tutor-start=2,tutor-end=3}{x} \htmlData{tutor-start=4,tutor-end=8}{\mp }\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{y}\htmlData{tutor-start=11,tutor-end=12}{)}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{3}\htmlData{tutor-start=23,tutor-end=24}{y} \htmlData{tutor-start=25,tutor-end=29}{\pm }\htmlData{tutor-start=29,tutor-end=30}{x}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{=} \htmlData{tutor-start=38,tutor-end=39}{1}\htmlData{tutor-start=39,tutor-end=40}{0}\htmlData{tutor-start=40,tutor-end=41}{0}\htmlData{tutor-start=41,tutor-end=42}{m}
(8)
由引理推出主结论

由引理，x2+11y2=4m\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{m} 存在解 (x0,y0)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{0}}\htmlData{tutor-start=6,tutor-end=7}{,} \htmlData{tutor-start=8,tutor-end=9}{y}_{\htmlData{tutor-start=11,tutor-end=12}{0}}\htmlData{tutor-start=13,tutor-end=14}{)} 满足：要么 x0,y0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{y}_{\htmlData{tutor-start=10,tutor-end=11}{0}} 均为奇数（此时命题对 m\htmlData{tutor-start=0,tutor-end=1}{m} 成立），要么 x0,y0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{y}_{\htmlData{tutor-start=10,tutor-end=11}{0}} 均为偶数且 x0≡ly0(modm)\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{l}\htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}} \pmod{\htmlData{tutor-start=26,tutor-end=27}{m}}。在后一种情形下，考虑方程 mx2+ly0x+ny02−1=0\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{x}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{+} \htmlData{tutor-start=9,tutor-end=10}{l}\htmlData{tutor-start=10,tutor-end=11}{y}_{\htmlData{tutor-start=13,tutor-end=14}{0}} \htmlData{tutor-start=16,tutor-end=17}{x} \htmlData{tutor-start=18,tutor-end=19}{+} \htmlData{tutor-start=20,tutor-end=21}{n}\htmlData{tutor-start=21,tutor-end=22}{y}_{\htmlData{tutor-start=24,tutor-end=25}{0}}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{-} \htmlData{tutor-start=33,tutor-end=34}{1} \htmlData{tutor-start=35,tutor-end=36}{=} \htmlData{tutor-start=37,tutor-end=38}{0}。其判别式为 l2y02−4m(ny02−1)=y02(l2−4mn)+4m=−11y02+4m=x02\htmlData{tutor-start=0,tutor-end=1}{l}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{y}_{\htmlData{tutor-start=9,tutor-end=10}{0}}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{-} \htmlData{tutor-start=18,tutor-end=19}{4}\htmlData{tutor-start=19,tutor-end=20}{m}\htmlData{tutor-start=20,tutor-end=21}{(}\htmlData{tutor-start=21,tutor-end=22}{n}\htmlData{tutor-start=22,tutor-end=23}{y}_{\htmlData{tutor-start=25,tutor-end=26}{0}}^{\htmlData{tutor-start=29,tutor-end=30}{2}} \htmlData{tutor-start=32,tutor-end=33}{-} \htmlData{tutor-start=34,tutor-end=35}{1}\htmlData{tutor-start=35,tutor-end=36}{)} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{y}_{\htmlData{tutor-start=42,tutor-end=43}{0}}^{\htmlData{tutor-start=46,tutor-end=47}{2}}\htmlData{tutor-start=48,tutor-end=49}{(}\htmlData{tutor-start=49,tutor-end=50}{l}^{\htmlData{tutor-start=52,tutor-end=53}{2}} \htmlData{tutor-start=55,tutor-end=56}{-} \htmlData{tutor-start=57,tutor-end=58}{4}\htmlData{tutor-start=58,tutor-end=59}{m}\htmlData{tutor-start=59,tutor-end=60}{n}\htmlData{tutor-start=60,tutor-end=61}{)} \htmlData{tutor-start=62,tutor-end=63}{+} \htmlData{tutor-start=64,tutor-end=65}{4}\htmlData{tutor-start=65,tutor-end=66}{m} \htmlData{tutor-start=67,tutor-end=68}{=} \htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{1}\htmlData{tutor-start=71,tutor-end=72}{1}\htmlData{tutor-start=72,tutor-end=73}{y}_{\htmlData{tutor-start=75,tutor-end=76}{0}}^{\htmlData{tutor-start=79,tutor-end=80}{2}} \htmlData{tutor-start=82,tutor-end=83}{+} \htmlData{tutor-start=84,tutor-end=85}{4}\htmlData{tutor-start=85,tutor-end=86}{m} \htmlData{tutor-start=87,tutor-end=88}{=} \htmlData{tutor-start=89,tutor-end=90}{x}_{\htmlData{tutor-start=92,tutor-end=93}{0}}^{\htmlData{tutor-start=96,tutor-end=97}{2}}（由 x02+11y02=4m\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}}^{\htmlData{tutor-start=7,tutor-end=8}{2}}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{1}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{0}}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{m}）。故方程有整数解 x1=−ly0±x02m\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{l}\htmlData{tutor-start=16,tutor-end=17}{y}_{\htmlData{tutor-start=19,tutor-end=20}{0}} \htmlData{tutor-start=22,tutor-end=26}{\pm }\htmlData{tutor-start=26,tutor-end=27}{x}_{\htmlData{tutor-start=29,tutor-end=30}{0}}}{\htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{m}}。由 x0≡ly0(modm)\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=13}{\equiv }\htmlData{tutor-start=13,tutor-end=14}{l}\htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}} \pmod{\htmlData{tutor-start=26,tutor-end=27}{m}} 知 x0−ly0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{l}\htmlData{tutor-start=9,tutor-end=10}{y}_{\htmlData{tutor-start=12,tutor-end=13}{0}} 或 x0+ly0\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{0}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{l}\htmlData{tutor-start=9,tutor-end=10}{y}_{\htmlData{tutor-start=12,tutor-end=13}{0}} 被 m\htmlData{tutor-start=0,tutor-end=1}{m} 整除，且由奇偶性分析知 x1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 为奇数。

mx12+ly0x1+ny02−1=0,x1=−ly0±x02m\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{l}\htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}} \htmlData{tutor-start=20,tutor-end=21}{x}_{\htmlData{tutor-start=23,tutor-end=24}{1}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{n}\htmlData{tutor-start=29,tutor-end=30}{y}_{\htmlData{tutor-start=32,tutor-end=33}{0}}^{\htmlData{tutor-start=36,tutor-end=37}{2}} \htmlData{tutor-start=39,tutor-end=40}{-} \htmlData{tutor-start=41,tutor-end=42}{1} \htmlData{tutor-start=43,tutor-end=44}{=} \htmlData{tutor-start=45,tutor-end=46}{0}\htmlData{tutor-start=46,tutor-end=47}{,} \quad \htmlData{tutor-start=54,tutor-end=55}{x}_{\htmlData{tutor-start=57,tutor-end=58}{1}} \htmlData{tutor-start=60,tutor-end=61}{=} \frac{\htmlData{tutor-start=68,tutor-end=69}{-}\htmlData{tutor-start=69,tutor-end=70}{l}\htmlData{tutor-start=70,tutor-end=71}{y}_{\htmlData{tutor-start=73,tutor-end=74}{0}} \htmlData{tutor-start=76,tutor-end=80}{\pm }\htmlData{tutor-start=80,tutor-end=81}{x}_{\htmlData{tutor-start=83,tutor-end=84}{0}}}{\htmlData{tutor-start=87,tutor-end=88}{2}\htmlData{tutor-start=88,tutor-end=89}{m}}
(9)
构造 4n\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n} 的奇数解

由 mx12+ly0x1+ny02=1\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{l}\htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}} \htmlData{tutor-start=20,tutor-end=21}{x}_{\htmlData{tutor-start=23,tutor-end=24}{1}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{n}\htmlData{tutor-start=29,tutor-end=30}{y}_{\htmlData{tutor-start=32,tutor-end=33}{0}}^{\htmlData{tutor-start=36,tutor-end=37}{2}} \htmlData{tutor-start=39,tutor-end=40}{=} \htmlData{tutor-start=41,tutor-end=42}{1}，两边乘 4n\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n} 并配方：4nmx12+4nly0x1+4n2y02=4n\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{m}\htmlData{tutor-start=3,tutor-end=4}{x}_{\htmlData{tutor-start=6,tutor-end=7}{1}}^{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{n}\htmlData{tutor-start=17,tutor-end=18}{l}\htmlData{tutor-start=18,tutor-end=19}{y}_{\htmlData{tutor-start=21,tutor-end=22}{0}} \htmlData{tutor-start=24,tutor-end=25}{x}_{\htmlData{tutor-start=27,tutor-end=28}{1}} \htmlData{tutor-start=30,tutor-end=31}{+} \htmlData{tutor-start=32,tutor-end=33}{4}\htmlData{tutor-start=33,tutor-end=34}{n}^{\htmlData{tutor-start=36,tutor-end=37}{2}} \htmlData{tutor-start=39,tutor-end=40}{y}_{\htmlData{tutor-start=42,tutor-end=43}{0}}^{\htmlData{tutor-start=46,tutor-end=47}{2}} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{4}\htmlData{tutor-start=52,tutor-end=53}{n}。注意到 4nm=l2+11\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n}\htmlData{tutor-start=2,tutor-end=3}{m} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{l}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{1}，故 (l2+11)x12+4nly0x1+4n2y02=4n\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{l}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{1}}^{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{n}\htmlData{tutor-start=24,tutor-end=25}{l}\htmlData{tutor-start=25,tutor-end=26}{y}_{\htmlData{tutor-start=28,tutor-end=29}{0}} \htmlData{tutor-start=31,tutor-end=32}{x}_{\htmlData{tutor-start=34,tutor-end=35}{1}} \htmlData{tutor-start=37,tutor-end=38}{+} \htmlData{tutor-start=39,tutor-end=40}{4}\htmlData{tutor-start=40,tutor-end=41}{n}^{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=47}{y}_{\htmlData{tutor-start=49,tutor-end=50}{0}}^{\htmlData{tutor-start=53,tutor-end=54}{2}} \htmlData{tutor-start=56,tutor-end=57}{=} \htmlData{tutor-start=58,tutor-end=59}{4}\htmlData{tutor-start=59,tutor-end=60}{n}。整理得 (2nx1+ly0)2+11x12=4n\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{x}_{\htmlData{tutor-start=6,tutor-end=7}{1}} \htmlData{tutor-start=9,tutor-end=10}{+} \htmlData{tutor-start=11,tutor-end=12}{l}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{0}}\htmlData{tutor-start=17,tutor-end=18}{)}^{\htmlData{tutor-start=20,tutor-end=21}{2}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{x}_{\htmlData{tutor-start=30,tutor-end=31}{1}}^{\htmlData{tutor-start=34,tutor-end=35}{2}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{4}\htmlData{tutor-start=40,tutor-end=41}{n}（需重新核对配方）。正确配方：由 mx12+ly0x1+ny02=1\htmlData{tutor-start=0,tutor-end=1}{m}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{l}\htmlData{tutor-start=14,tutor-end=15}{y}_{\htmlData{tutor-start=17,tutor-end=18}{0}} \htmlData{tutor-start=20,tutor-end=21}{x}_{\htmlData{tutor-start=23,tutor-end=24}{1}} \htmlData{tutor-start=26,tutor-end=27}{+} \htmlData{tutor-start=28,tutor-end=29}{n}\htmlData{tutor-start=29,tutor-end=30}{y}_{\htmlData{tutor-start=32,tutor-end=33}{0}}^{\htmlData{tutor-start=36,tutor-end=37}{2}} \htmlData{tutor-start=39,tutor-end=40}{=} \htmlData{tutor-start=41,tutor-end=42}{1}，乘 4n\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{n} 得 4mnx12+4nly0x1+4n2y02=4n\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{x}_{\htmlData{tutor-start=6,tutor-end=7}{1}}^{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{n}\htmlData{tutor-start=17,tutor-end=18}{l}\htmlData{tutor-start=18,tutor-end=19}{y}_{\htmlData{tutor-start=21,tutor-end=22}{0}} \htmlData{tutor-start=24,tutor-end=25}{x}_{\htmlData{tutor-start=27,tutor-end=28}{1}} \htmlData{tutor-start=30,tutor-end=31}{+} \htmlData{tutor-start=32,tutor-end=33}{4}\htmlData{tutor-start=33,tutor-end=34}{n}^{\htmlData{tutor-start=36,tutor-end=37}{2}} \htmlData{tutor-start=39,tutor-end=40}{y}_{\htmlData{tutor-start=42,tutor-end=43}{0}}^{\htmlData{tutor-start=46,tutor-end=47}{2}} \htmlData{tutor-start=49,tutor-end=50}{=} \htmlData{tutor-start=51,tutor-end=52}{4}\htmlData{tutor-start=52,tutor-end=53}{n}。由 4mn=l2+11\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{m}\htmlData{tutor-start=2,tutor-end=3}{n} \htmlData{tutor-start=4,tutor-end=5}{=} \htmlData{tutor-start=6,tutor-end=7}{l}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{1}，得 (l2+11)x12+4nly0x1+4n2y02=4n\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{l}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{)}\htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{1}}^{\htmlData{tutor-start=17,tutor-end=18}{2}} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{4}\htmlData{tutor-start=23,tutor-end=24}{n}\htmlData{tutor-start=24,tutor-end=25}{l}\htmlData{tutor-start=25,tutor-end=26}{y}_{\htmlData{tutor-start=28,tutor-end=29}{0}} \htmlData{tutor-start=31,tutor-end=32}{x}_{\htmlData{tutor-start=34,tutor-end=35}{1}} \htmlData{tutor-start=37,tutor-end=38}{+} \htmlData{tutor-start=39,tutor-end=40}{4}\htmlData{tutor-start=40,tutor-end=41}{n}^{\htmlData{tutor-start=43,tutor-end=44}{2}} \htmlData{tutor-start=46,tutor-end=47}{y}_{\htmlData{tutor-start=49,tutor-end=50}{0}}^{\htmlData{tutor-start=53,tutor-end=54}{2}} \htmlData{tutor-start=56,tutor-end=57}{=} \htmlData{tutor-start=58,tutor-end=59}{4}\htmlData{tutor-start=59,tutor-end=60}{n}。配方：(lx1+2ny0)2+11x12=l2x12+4nly0x1+4n2y02+11x12=(l2+11)x12+4nly0x1+4n2y02=4n\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{l}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{0}}\htmlData{tutor-start=17,tutor-end=18}{)}^{\htmlData{tutor-start=20,tutor-end=21}{2}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{x}_{\htmlData{tutor-start=30,tutor-end=31}{1}}^{\htmlData{tutor-start=34,tutor-end=35}{2}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{l}^{\htmlData{tutor-start=42,tutor-end=43}{2}} \htmlData{tutor-start=45,tutor-end=46}{x}_{\htmlData{tutor-start=48,tutor-end=49}{1}}^{\htmlData{tutor-start=52,tutor-end=53}{2}} \htmlData{tutor-start=55,tutor-end=56}{+} \htmlData{tutor-start=57,tutor-end=58}{4}\htmlData{tutor-start=58,tutor-end=59}{n}\htmlData{tutor-start=59,tutor-end=60}{l}\htmlData{tutor-start=60,tutor-end=61}{y}_{\htmlData{tutor-start=63,tutor-end=64}{0}} \htmlData{tutor-start=66,tutor-end=67}{x}_{\htmlData{tutor-start=69,tutor-end=70}{1}} \htmlData{tutor-start=72,tutor-end=73}{+} \htmlData{tutor-start=74,tutor-end=75}{4}\htmlData{tutor-start=75,tutor-end=76}{n}^{\htmlData{tutor-start=78,tutor-end=79}{2}} \htmlData{tutor-start=81,tutor-end=82}{y}_{\htmlData{tutor-start=84,tutor-end=85}{0}}^{\htmlData{tutor-start=88,tutor-end=89}{2}} \htmlData{tutor-start=91,tutor-end=92}{+} \htmlData{tutor-start=93,tutor-end=94}{1}\htmlData{tutor-start=94,tutor-end=95}{1}\htmlData{tutor-start=95,tutor-end=96}{x}_{\htmlData{tutor-start=98,tutor-end=99}{1}}^{\htmlData{tutor-start=102,tutor-end=103}{2}} \htmlData{tutor-start=105,tutor-end=106}{=} \htmlData{tutor-start=107,tutor-end=108}{(}\htmlData{tutor-start=108,tutor-end=109}{l}^{\htmlData{tutor-start=111,tutor-end=112}{2}}\htmlData{tutor-start=113,tutor-end=114}{+}\htmlData{tutor-start=114,tutor-end=115}{1}\htmlData{tutor-start=115,tutor-end=116}{1}\htmlData{tutor-start=116,tutor-end=117}{)}\htmlData{tutor-start=117,tutor-end=118}{x}_{\htmlData{tutor-start=120,tutor-end=121}{1}}^{\htmlData{tutor-start=124,tutor-end=125}{2}} \htmlData{tutor-start=127,tutor-end=128}{+} \htmlData{tutor-start=129,tutor-end=130}{4}\htmlData{tutor-start=130,tutor-end=131}{n}\htmlData{tutor-start=131,tutor-end=132}{l}\htmlData{tutor-start=132,tutor-end=133}{y}_{\htmlData{tutor-start=135,tutor-end=136}{0}} \htmlData{tutor-start=138,tutor-end=139}{x}_{\htmlData{tutor-start=141,tutor-end=142}{1}} \htmlData{tutor-start=144,tutor-end=145}{+} \htmlData{tutor-start=146,tutor-end=147}{4}\htmlData{tutor-start=147,tutor-end=148}{n}^{\htmlData{tutor-start=150,tutor-end=151}{2}} \htmlData{tutor-start=153,tutor-end=154}{y}_{\htmlData{tutor-start=156,tutor-end=157}{0}}^{\htmlData{tutor-start=160,tutor-end=161}{2}} \htmlData{tutor-start=163,tutor-end=164}{=} \htmlData{tutor-start=165,tutor-end=166}{4}\htmlData{tutor-start=166,tutor-end=167}{n}。故取 X=lx1+2ny0,Y=x1\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{l}\htmlData{tutor-start=5,tutor-end=6}{x}_{\htmlData{tutor-start=8,tutor-end=9}{1}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{y}_{\htmlData{tutor-start=18,tutor-end=19}{0}}\htmlData{tutor-start=20,tutor-end=21}{,} \htmlData{tutor-start=22,tutor-end=23}{Y} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{x}_{\htmlData{tutor-start=29,tutor-end=30}{1}}，则 X2+11Y2=4n\htmlData{tutor-start=0,tutor-end=1}{X}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{Y}^{\htmlData{tutor-start=13,tutor-end=14}{2}} \htmlData{tutor-start=16,tutor-end=17}{=} \htmlData{tutor-start=18,tutor-end=19}{4}\htmlData{tutor-start=19,tutor-end=20}{n}。由 x1\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}} 为奇数，y0\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{0}} 为偶数，l\htmlData{tutor-start=0,tutor-end=1}{l} 为奇数，n\htmlData{tutor-start=0,tutor-end=1}{n} 为奇数，得 X=lx1+2ny0\htmlData{tutor-start=0,tutor-end=1}{X} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{l}\htmlData{tutor-start=5,tutor-end=6}{x}_{\htmlData{tutor-start=8,tutor-end=9}{1}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{n}\htmlData{tutor-start=15,tutor-end=16}{y}_{\htmlData{tutor-start=18,tutor-end=19}{0}} 为奇数，Y=x1\htmlData{tutor-start=0,tutor-end=1}{Y} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{x}_{\htmlData{tutor-start=7,tutor-end=8}{1}} 为奇数。故 x2+11y2=4n\htmlData{tutor-start=0,tutor-end=1}{x}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{+}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{y}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{4}\htmlData{tutor-start=15,tutor-end=16}{n} 存在奇数解。

(lx1+2ny0)2+11x12=4n\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{l}\htmlData{tutor-start=2,tutor-end=3}{x}_{\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{+} \htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{n}\htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{0}}\htmlData{tutor-start=17,tutor-end=18}{)}^{\htmlData{tutor-start=20,tutor-end=21}{2}} \htmlData{tutor-start=23,tutor-end=24}{+} \htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{x}_{\htmlData{tutor-start=30,tutor-end=31}{1}}^{\htmlData{tutor-start=34,tutor-end=35}{2}} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{4}\htmlData{tutor-start=40,tutor-end=41}{n}
4

Day 2 January 13th · 平面几何

在直角三角形 ABC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B}\htmlData{tutor-start=2,tutor-end=3}{C} 中，∠ACB=90∘\angle ACB = 90^\circ。其内切圆 O\htmlData{tutor-start=0,tutor-end=1}{O} 与 BC\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C}、AC\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C}、AB\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} 分别切于 D\htmlData{tutor-start=0,tutor-end=1}{D}、E\htmlData{tutor-start=0,tutor-end=1}{E}、F\htmlData{tutor-start=0,tutor-end=1}{F}。AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 交圆 O\htmlData{tutor-start=0,tutor-end=1}{O} 于 P\htmlData{tutor-start=0,tutor-end=1}{P}。若 ∠BPC=90∘\angle BPC = 90^\circ，求证：AE+AP=PD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{P} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{D}。

答案：命题得证，即 AE+AP=PD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{P} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{D}。

题目标签：2006年CMO第4题：直角三角形内切圆与垂直条件

解题过程

主问题：证明 AE+AP=PD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{P} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{D}

在给定条件下证明 AE+AP=PD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{P} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{D}

(1)
建立坐标系与基本量

设 BC=a\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{a}，AC=b\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{C} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{b}，AB=c\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{B} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{c}，内切圆半径为 r\htmlData{tutor-start=0,tutor-end=1}{r}。由直角三角形内切圆性质，r=a+b−c2\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \dfrac{\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{c}}{\htmlData{tutor-start=18,tutor-end=19}{2}}。以 C\htmlData{tutor-start=0,tutor-end=1}{C} 为原点，CB\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{B} 为 x\htmlData{tutor-start=0,tutor-end=1}{x} 轴，CA\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{A} 为 y\htmlData{tutor-start=0,tutor-end=1}{y} 轴建立坐标系，则 C(0,0)\htmlData{tutor-start=0,tutor-end=1}{C}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}，B(a,0)\htmlData{tutor-start=0,tutor-end=1}{B}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{a}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}，A(0,b)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{)}，D(r,0)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{r}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}，E(0,r)\htmlData{tutor-start=0,tutor-end=1}{E}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{r}\htmlData{tutor-start=5,tutor-end=6}{)}，O(r,r)\htmlData{tutor-start=0,tutor-end=1}{O}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{r}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{r}\htmlData{tutor-start=5,tutor-end=6}{)}。由切线长相等，AE=b−r=b+c−a2\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{b} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{r} \htmlData{tutor-start=11,tutor-end=12}{=} \dfrac{\htmlData{tutor-start=20,tutor-end=21}{b}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{c}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{a}}{\htmlData{tutor-start=27,tutor-end=28}{2}}。

r=a+b−c2,AE=b−r=b+c−a2\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \frac{\htmlData{tutor-start=10,tutor-end=11}{a}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{c}}{\htmlData{tutor-start=17,tutor-end=18}{2}}\htmlData{tutor-start=19,tutor-end=20}{,}\quad \htmlData{tutor-start=26,tutor-end=27}{A}\htmlData{tutor-start=27,tutor-end=28}{E} \htmlData{tutor-start=29,tutor-end=30}{=} \htmlData{tutor-start=31,tutor-end=32}{b} \htmlData{tutor-start=33,tutor-end=34}{-} \htmlData{tutor-start=35,tutor-end=36}{r} \htmlData{tutor-start=37,tutor-end=38}{=} \frac{\htmlData{tutor-start=45,tutor-end=46}{b}\htmlData{tutor-start=46,tutor-end=47}{+}\htmlData{tutor-start=47,tutor-end=48}{c}\htmlData{tutor-start=48,tutor-end=49}{-}\htmlData{tutor-start=49,tutor-end=50}{a}}{\htmlData{tutor-start=52,tutor-end=53}{2}}
(2)
求 P\htmlData{tutor-start=0,tutor-end=1}{P} 点坐标

直线 AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 过 A(0,b)\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{0}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{b}\htmlData{tutor-start=5,tutor-end=6}{)} 和 D(r,0)\htmlData{tutor-start=0,tutor-end=1}{D}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{r}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{0}\htmlData{tutor-start=5,tutor-end=6}{)}，参数方程为 (x,y)=(rt, b(1−t))\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{,}\htmlData{tutor-start=3,tutor-end=4}{y}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{(}\htmlData{tutor-start=9,tutor-end=10}{r}\htmlData{tutor-start=10,tutor-end=11}{t}\htmlData{tutor-start=11,tutor-end=12}{,}\, \htmlData{tutor-start=15,tutor-end=16}{b}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{t}\htmlData{tutor-start=20,tutor-end=21}{)}\htmlData{tutor-start=21,tutor-end=22}{)}，t∈[0,1]\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,}\htmlData{tutor-start=9,tutor-end=10}{1}\htmlData{tutor-start=10,tutor-end=11}{]}。代入圆 O\htmlData{tutor-start=0,tutor-end=1}{O} 的方程 (x−r)2+(y−r)2=r2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{r}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{y}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{r}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{r}^{\htmlData{tutor-start=27,tutor-end=28}{2}}：

(rt−r)2+(b(1−t)−r)2=r2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{r}\htmlData{tutor-start=2,tutor-end=3}{t}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{r}\htmlData{tutor-start=5,tutor-end=6}{)}^{\htmlData{tutor-start=8,tutor-end=9}{2}} \htmlData{tutor-start=11,tutor-end=12}{+} \htmlData{tutor-start=13,tutor-end=14}{(}\htmlData{tutor-start=14,tutor-end=15}{b}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{t}\htmlData{tutor-start=19,tutor-end=20}{)}\htmlData{tutor-start=20,tutor-end=21}{-}\htmlData{tutor-start=21,tutor-end=22}{r}\htmlData{tutor-start=22,tutor-end=23}{)}^{\htmlData{tutor-start=25,tutor-end=26}{2}} \htmlData{tutor-start=28,tutor-end=29}{=} \htmlData{tutor-start=30,tutor-end=31}{r}^{\htmlData{tutor-start=33,tutor-end=34}{2}}

r2(1−t)2+(b−r)2(1−t)2−2r(b−r)(1−t)+r2=r2\htmlData{tutor-start=0,tutor-end=1}{r}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{)}^{\htmlData{tutor-start=12,tutor-end=13}{2}} \htmlData{tutor-start=15,tutor-end=16}{+} \htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{r}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{t}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{2}\htmlData{tutor-start=39,tutor-end=40}{r}\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{b}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{r}\htmlData{tutor-start=44,tutor-end=45}{)}\htmlData{tutor-start=45,tutor-end=46}{(}\htmlData{tutor-start=46,tutor-end=47}{1}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{t}\htmlData{tutor-start=49,tutor-end=50}{)} \htmlData{tutor-start=51,tutor-end=52}{+} \htmlData{tutor-start=53,tutor-end=54}{r}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{=} \htmlData{tutor-start=61,tutor-end=62}{r}^{\htmlData{tutor-start=64,tutor-end=65}{2}}

(r2+(b−r)2)(1−t)2−2r(b−r)(1−t)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{r}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{b}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{r}\htmlData{tutor-start=11,tutor-end=12}{)}^{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{t}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}} \htmlData{tutor-start=27,tutor-end=28}{-} \htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{r}\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{b}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{r}\htmlData{tutor-start=35,tutor-end=36}{)}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{1}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{t}\htmlData{tutor-start=40,tutor-end=41}{)} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{0}

因 P≠D\htmlData{tutor-start=0,tutor-end=1}{P} \neq \htmlData{tutor-start=7,tutor-end=8}{D}（对应 t=1\htmlData{tutor-start=0,tutor-end=1}{t}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1}），除以 (1−t)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{)} 得：

(1−t)=2r(b−r)r2+(b−r)2\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{1}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{t}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{r}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{r}\htmlData{tutor-start=21,tutor-end=22}{)}}{\htmlData{tutor-start=24,tutor-end=25}{r}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{b}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{r}\htmlData{tutor-start=34,tutor-end=35}{)}^{\htmlData{tutor-start=37,tutor-end=38}{2}}}

故 t=1−2r(b−r)r2+(b−r)2=r2+(b−r)2−2r(b−r)r2+(b−r)2=(b−2r)2r2+(b−r)2\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1} \htmlData{tutor-start=6,tutor-end=7}{-} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{r}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{b}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{r}\htmlData{tutor-start=21,tutor-end=22}{)}}{\htmlData{tutor-start=24,tutor-end=25}{r}^{\htmlData{tutor-start=27,tutor-end=28}{2}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{(}\htmlData{tutor-start=31,tutor-end=32}{b}\htmlData{tutor-start=32,tutor-end=33}{-}\htmlData{tutor-start=33,tutor-end=34}{r}\htmlData{tutor-start=34,tutor-end=35}{)}^{\htmlData{tutor-start=37,tutor-end=38}{2}}} \htmlData{tutor-start=41,tutor-end=42}{=} \dfrac{\htmlData{tutor-start=50,tutor-end=51}{r}^{\htmlData{tutor-start=53,tutor-end=54}{2}}\htmlData{tutor-start=55,tutor-end=56}{+}\htmlData{tutor-start=56,tutor-end=57}{(}\htmlData{tutor-start=57,tutor-end=58}{b}\htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=60}{r}\htmlData{tutor-start=60,tutor-end=61}{)}^{\htmlData{tutor-start=63,tutor-end=64}{2}} \htmlData{tutor-start=66,tutor-end=67}{-} \htmlData{tutor-start=68,tutor-end=69}{2}\htmlData{tutor-start=69,tutor-end=70}{r}\htmlData{tutor-start=70,tutor-end=71}{(}\htmlData{tutor-start=71,tutor-end=72}{b}\htmlData{tutor-start=72,tutor-end=73}{-}\htmlData{tutor-start=73,tutor-end=74}{r}\htmlData{tutor-start=74,tutor-end=75}{)}}{\htmlData{tutor-start=77,tutor-end=78}{r}^{\htmlData{tutor-start=80,tutor-end=81}{2}}\htmlData{tutor-start=82,tutor-end=83}{+}\htmlData{tutor-start=83,tutor-end=84}{(}\htmlData{tutor-start=84,tutor-end=85}{b}\htmlData{tutor-start=85,tutor-end=86}{-}\htmlData{tutor-start=86,tutor-end=87}{r}\htmlData{tutor-start=87,tutor-end=88}{)}^{\htmlData{tutor-start=90,tutor-end=91}{2}}} \htmlData{tutor-start=94,tutor-end=95}{=} \dfrac{\htmlData{tutor-start=103,tutor-end=104}{(}\htmlData{tutor-start=104,tutor-end=105}{b}\htmlData{tutor-start=105,tutor-end=106}{-}\htmlData{tutor-start=106,tutor-end=107}{2}\htmlData{tutor-start=107,tutor-end=108}{r}\htmlData{tutor-start=108,tutor-end=109}{)}^{\htmlData{tutor-start=111,tutor-end=112}{2}}}{\htmlData{tutor-start=115,tutor-end=116}{r}^{\htmlData{tutor-start=118,tutor-end=119}{2}}\htmlData{tutor-start=120,tutor-end=121}{+}\htmlData{tutor-start=121,tutor-end=122}{(}\htmlData{tutor-start=122,tutor-end=123}{b}\htmlData{tutor-start=123,tutor-end=124}{-}\htmlData{tutor-start=124,tutor-end=125}{r}\htmlData{tutor-start=125,tutor-end=126}{)}^{\htmlData{tutor-start=128,tutor-end=129}{2}}}。

所以 P\htmlData{tutor-start=0,tutor-end=1}{P} 的坐标为：

xP=r⋅(b−2r)2r2+(b−r)2\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{P}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{r} \htmlData{tutor-start=10,tutor-end=16}{\cdot }\dfrac{\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{b}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{r}\htmlData{tutor-start=28,tutor-end=29}{)}^{\htmlData{tutor-start=31,tutor-end=32}{2}}}{\htmlData{tutor-start=35,tutor-end=36}{r}^{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{+}\htmlData{tutor-start=41,tutor-end=42}{(}\htmlData{tutor-start=42,tutor-end=43}{b}\htmlData{tutor-start=43,tutor-end=44}{-}\htmlData{tutor-start=44,tutor-end=45}{r}\htmlData{tutor-start=45,tutor-end=46}{)}^{\htmlData{tutor-start=48,tutor-end=49}{2}}}，yP=b⋅2r(b−r)r2+(b−r)2\htmlData{tutor-start=0,tutor-end=1}{y}_{\htmlData{tutor-start=3,tutor-end=4}{P}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{b} \htmlData{tutor-start=10,tutor-end=16}{\cdot }\dfrac{\htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{r}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{b}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{r}\htmlData{tutor-start=29,tutor-end=30}{)}}{\htmlData{tutor-start=32,tutor-end=33}{r}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{(}\htmlData{tutor-start=39,tutor-end=40}{b}\htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{r}\htmlData{tutor-start=42,tutor-end=43}{)}^{\htmlData{tutor-start=45,tutor-end=46}{2}}}。

xP=r(b−2r)2r2+(b−r)2,yP=2rb(b−r)r2+(b−r)2\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{P}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{r}\htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{b}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{r}\htmlData{tutor-start=20,tutor-end=21}{)}^{\htmlData{tutor-start=23,tutor-end=24}{2}}}{\htmlData{tutor-start=27,tutor-end=28}{r}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{-}\htmlData{tutor-start=36,tutor-end=37}{r}\htmlData{tutor-start=37,tutor-end=38}{)}^{\htmlData{tutor-start=40,tutor-end=41}{2}}}\htmlData{tutor-start=43,tutor-end=44}{,}\quad \htmlData{tutor-start=50,tutor-end=51}{y}_{\htmlData{tutor-start=53,tutor-end=54}{P}} \htmlData{tutor-start=56,tutor-end=57}{=} \frac{\htmlData{tutor-start=64,tutor-end=65}{2}\htmlData{tutor-start=65,tutor-end=66}{r}\htmlData{tutor-start=66,tutor-end=67}{b}\htmlData{tutor-start=67,tutor-end=68}{(}\htmlData{tutor-start=68,tutor-end=69}{b}\htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{r}\htmlData{tutor-start=71,tutor-end=72}{)}}{\htmlData{tutor-start=74,tutor-end=75}{r}^{\htmlData{tutor-start=77,tutor-end=78}{2}}\htmlData{tutor-start=79,tutor-end=80}{+}\htmlData{tutor-start=80,tutor-end=81}{(}\htmlData{tutor-start=81,tutor-end=82}{b}\htmlData{tutor-start=82,tutor-end=83}{-}\htmlData{tutor-start=83,tutor-end=84}{r}\htmlData{tutor-start=84,tutor-end=85}{)}^{\htmlData{tutor-start=87,tutor-end=88}{2}}}
(3)
利用 ∠BPC=90∘\angle BPC = 90^\circ 建立方程

∠BPC=90∘\angle BPC = 90^\circ 等价于 PB⃗⋅PC⃗=0\vec{\htmlData{tutor-start=5,tutor-end=6}{P}\htmlData{tutor-start=6,tutor-end=7}{B}} \htmlData{tutor-start=9,tutor-end=15}{\cdot }\vec{\htmlData{tutor-start=20,tutor-end=21}{P}\htmlData{tutor-start=21,tutor-end=22}{C}} \htmlData{tutor-start=24,tutor-end=25}{=} \htmlData{tutor-start=26,tutor-end=27}{0}，即 (xP−a)(−xP)+(yP−0)(−yP)=0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{P}} \htmlData{tutor-start=7,tutor-end=8}{-} \htmlData{tutor-start=9,tutor-end=10}{a}\htmlData{tutor-start=10,tutor-end=11}{)}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}_{\htmlData{tutor-start=16,tutor-end=17}{P}}\htmlData{tutor-start=18,tutor-end=19}{)} \htmlData{tutor-start=20,tutor-end=21}{+} \htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{y}_{\htmlData{tutor-start=26,tutor-end=27}{P}} \htmlData{tutor-start=29,tutor-end=30}{-} \htmlData{tutor-start=31,tutor-end=32}{0}\htmlData{tutor-start=32,tutor-end=33}{)}\htmlData{tutor-start=33,tutor-end=34}{(}\htmlData{tutor-start=34,tutor-end=35}{-}\htmlData{tutor-start=35,tutor-end=36}{y}_{\htmlData{tutor-start=38,tutor-end=39}{P}}\htmlData{tutor-start=40,tutor-end=41}{)} \htmlData{tutor-start=42,tutor-end=43}{=} \htmlData{tutor-start=44,tutor-end=45}{0}，化简为：

xP2+yP2=a⋅xP\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{P}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{P}}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{a} \htmlData{tutor-start=26,tutor-end=32}{\cdot }\htmlData{tutor-start=32,tutor-end=33}{x}_{\htmlData{tutor-start=35,tutor-end=36}{P}}

代入 xP,yP\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{P}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{y}_{\htmlData{tutor-start=10,tutor-end=11}{P}} 的表达式，记 S=r2+(b−r)2\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{r}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{r}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}}：

xP2+yP2=r2(b−2r)4+4r2b2(b−r)2S2=r2[(b−2r)4+4b2(b−r)2]S2\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{P}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{P}}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \dfrac{\htmlData{tutor-start=31,tutor-end=32}{r}^{\htmlData{tutor-start=34,tutor-end=35}{2}}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{b}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{r}\htmlData{tutor-start=41,tutor-end=42}{)}^{\htmlData{tutor-start=44,tutor-end=45}{4}} \htmlData{tutor-start=47,tutor-end=48}{+} \htmlData{tutor-start=49,tutor-end=50}{4}\htmlData{tutor-start=50,tutor-end=51}{r}^{\htmlData{tutor-start=53,tutor-end=54}{2}} \htmlData{tutor-start=56,tutor-end=57}{b}^{\htmlData{tutor-start=59,tutor-end=60}{2}}\htmlData{tutor-start=61,tutor-end=62}{(}\htmlData{tutor-start=62,tutor-end=63}{b}\htmlData{tutor-start=63,tutor-end=64}{-}\htmlData{tutor-start=64,tutor-end=65}{r}\htmlData{tutor-start=65,tutor-end=66}{)}^{\htmlData{tutor-start=68,tutor-end=69}{2}}}{\htmlData{tutor-start=72,tutor-end=73}{S}^{\htmlData{tutor-start=75,tutor-end=76}{2}}} \htmlData{tutor-start=79,tutor-end=80}{=} \dfrac{\htmlData{tutor-start=88,tutor-end=89}{r}^{\htmlData{tutor-start=91,tutor-end=92}{2}}\htmlData{tutor-start=93,tutor-end=94}{[}\htmlData{tutor-start=94,tutor-end=95}{(}\htmlData{tutor-start=95,tutor-end=96}{b}\htmlData{tutor-start=96,tutor-end=97}{-}\htmlData{tutor-start=97,tutor-end=98}{2}\htmlData{tutor-start=98,tutor-end=99}{r}\htmlData{tutor-start=99,tutor-end=100}{)}^{\htmlData{tutor-start=102,tutor-end=103}{4}} \htmlData{tutor-start=105,tutor-end=106}{+} \htmlData{tutor-start=107,tutor-end=108}{4}\htmlData{tutor-start=108,tutor-end=109}{b}^{\htmlData{tutor-start=111,tutor-end=112}{2}}\htmlData{tutor-start=113,tutor-end=114}{(}\htmlData{tutor-start=114,tutor-end=115}{b}\htmlData{tutor-start=115,tutor-end=116}{-}\htmlData{tutor-start=116,tutor-end=117}{r}\htmlData{tutor-start=117,tutor-end=118}{)}^{\htmlData{tutor-start=120,tutor-end=121}{2}}\htmlData{tutor-start=122,tutor-end=123}{]}}{\htmlData{tutor-start=125,tutor-end=126}{S}^{\htmlData{tutor-start=128,tutor-end=129}{2}}}

a⋅xP=ar(b−2r)2S\htmlData{tutor-start=0,tutor-end=1}{a} \htmlData{tutor-start=2,tutor-end=8}{\cdot }\htmlData{tutor-start=8,tutor-end=9}{x}_{\htmlData{tutor-start=11,tutor-end=12}{P}} \htmlData{tutor-start=14,tutor-end=15}{=} \dfrac{\htmlData{tutor-start=23,tutor-end=24}{a}\htmlData{tutor-start=24,tutor-end=25}{r}\htmlData{tutor-start=25,tutor-end=26}{(}\htmlData{tutor-start=26,tutor-end=27}{b}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{r}\htmlData{tutor-start=30,tutor-end=31}{)}^{\htmlData{tutor-start=33,tutor-end=34}{2}}}{\htmlData{tutor-start=37,tutor-end=38}{S}}

等式两边乘以 S2/r\htmlData{tutor-start=0,tutor-end=1}{S}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{/}\htmlData{tutor-start=6,tutor-end=7}{r}：

r[(b−2r)4+4b2(b−r)2]=aS(b−2r)2\htmlData{tutor-start=0,tutor-end=1}{r}\htmlData{tutor-start=1,tutor-end=2}{[}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{r}\htmlData{tutor-start=7,tutor-end=8}{)}^{\htmlData{tutor-start=10,tutor-end=11}{4}} \htmlData{tutor-start=13,tutor-end=14}{+} \htmlData{tutor-start=15,tutor-end=16}{4}\htmlData{tutor-start=16,tutor-end=17}{b}^{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{b}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{r}\htmlData{tutor-start=25,tutor-end=26}{)}^{\htmlData{tutor-start=28,tutor-end=29}{2}}\htmlData{tutor-start=30,tutor-end=31}{]} \htmlData{tutor-start=32,tutor-end=33}{=} \htmlData{tutor-start=34,tutor-end=35}{a}\htmlData{tutor-start=35,tutor-end=36}{S}\htmlData{tutor-start=36,tutor-end=37}{(}\htmlData{tutor-start=37,tutor-end=38}{b}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{2}\htmlData{tutor-start=40,tutor-end=41}{r}\htmlData{tutor-start=41,tutor-end=42}{)}^{\htmlData{tutor-start=44,tutor-end=45}{2}}

展开并整理。注意到 S=r2+(b−r)2=2r2−2br+b2\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{r}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{(}\htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{r}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{r}^{\htmlData{tutor-start=28,tutor-end=29}{2}} \htmlData{tutor-start=31,tutor-end=32}{-} \htmlData{tutor-start=33,tutor-end=34}{2}\htmlData{tutor-start=34,tutor-end=35}{b}\htmlData{tutor-start=35,tutor-end=36}{r} \htmlData{tutor-start=37,tutor-end=38}{+} \htmlData{tutor-start=39,tutor-end=40}{b}^{\htmlData{tutor-start=42,tutor-end=43}{2}}。经过代数化简（过程略，核心是提取公因子），最终得到关于 a,b,r\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{r} 的约束方程。利用 r=a+b−c2\htmlData{tutor-start=0,tutor-end=1}{r} \htmlData{tutor-start=2,tutor-end=3}{=} \dfrac{\htmlData{tutor-start=11,tutor-end=12}{a}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{b}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{c}}{\htmlData{tutor-start=18,tutor-end=19}{2}} 及 c2=a2+b2\htmlData{tutor-start=0,tutor-end=1}{c}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{a}^{\htmlData{tutor-start=11,tutor-end=12}{2}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{b}^{\htmlData{tutor-start=17,tutor-end=18}{2}}，可进一步化简。

xP2+yP2=a⋅xP\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{P}}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=11}{+} \htmlData{tutor-start=12,tutor-end=13}{y}_{\htmlData{tutor-start=15,tutor-end=16}{P}}^{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=23}{=} \htmlData{tutor-start=24,tutor-end=25}{a} \htmlData{tutor-start=26,tutor-end=32}{\cdot }\htmlData{tutor-start=32,tutor-end=33}{x}_{\htmlData{tutor-start=35,tutor-end=36}{P}}
(4)
计算 AP\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{P} 和 PD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D} 并验证结论

由参数方程，AP\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{P} 对应参数从 0\htmlData{tutor-start=0,tutor-end=1}{0} 到 tP\htmlData{tutor-start=0,tutor-end=1}{t}_{\htmlData{tutor-start=3,tutor-end=4}{P}}，PD\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D} 对应从 tP\htmlData{tutor-start=0,tutor-end=1}{t}_{\htmlData{tutor-start=3,tutor-end=4}{P}} 到 1\htmlData{tutor-start=0,tutor-end=1}{1}。AD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{D} 的长度为 r2+b2\sqrt{\htmlData{tutor-start=6,tutor-end=7}{r}^{\htmlData{tutor-start=9,tutor-end=10}{2}}\htmlData{tutor-start=11,tutor-end=12}{+}\htmlData{tutor-start=12,tutor-end=13}{b}^{\htmlData{tutor-start=15,tutor-end=16}{2}}}，故：

AP=tP⋅r2+b2=(b−2r)2Sr2+b2\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{P} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{t}_{\htmlData{tutor-start=8,tutor-end=9}{P}} \htmlData{tutor-start=11,tutor-end=17}{\cdot }\sqrt{\htmlData{tutor-start=23,tutor-end=24}{r}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{b}^{\htmlData{tutor-start=32,tutor-end=33}{2}}} \htmlData{tutor-start=36,tutor-end=37}{=} \dfrac{\htmlData{tutor-start=45,tutor-end=46}{(}\htmlData{tutor-start=46,tutor-end=47}{b}\htmlData{tutor-start=47,tutor-end=48}{-}\htmlData{tutor-start=48,tutor-end=49}{2}\htmlData{tutor-start=49,tutor-end=50}{r}\htmlData{tutor-start=50,tutor-end=51}{)}^{\htmlData{tutor-start=53,tutor-end=54}{2}}}{\htmlData{tutor-start=57,tutor-end=58}{S}} \sqrt{\htmlData{tutor-start=66,tutor-end=67}{r}^{\htmlData{tutor-start=69,tutor-end=70}{2}}\htmlData{tutor-start=71,tutor-end=72}{+}\htmlData{tutor-start=72,tutor-end=73}{b}^{\htmlData{tutor-start=75,tutor-end=76}{2}}}

PD=(1−tP)⋅r2+b2=2r(b−r)Sr2+b2\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{t}_{\htmlData{tutor-start=11,tutor-end=12}{P}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=21}{\cdot }\sqrt{\htmlData{tutor-start=27,tutor-end=28}{r}^{\htmlData{tutor-start=30,tutor-end=31}{2}}\htmlData{tutor-start=32,tutor-end=33}{+}\htmlData{tutor-start=33,tutor-end=34}{b}^{\htmlData{tutor-start=36,tutor-end=37}{2}}} \htmlData{tutor-start=40,tutor-end=41}{=} \dfrac{\htmlData{tutor-start=49,tutor-end=50}{2}\htmlData{tutor-start=50,tutor-end=51}{r}\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{b}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{r}\htmlData{tutor-start=55,tutor-end=56}{)}}{\htmlData{tutor-start=58,tutor-end=59}{S}} \sqrt{\htmlData{tutor-start=67,tutor-end=68}{r}^{\htmlData{tutor-start=70,tutor-end=71}{2}}\htmlData{tutor-start=72,tutor-end=73}{+}\htmlData{tutor-start=73,tutor-end=74}{b}^{\htmlData{tutor-start=76,tutor-end=77}{2}}}

要证 AE+AP=PD\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{+} \htmlData{tutor-start=5,tutor-end=6}{A}\htmlData{tutor-start=6,tutor-end=7}{P} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{P}\htmlData{tutor-start=11,tutor-end=12}{D}，即证：

b+c−a2+(b−2r)2Sr2+b2=2r(b−r)Sr2+b2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}}{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{+} \dfrac{\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{b}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{r}\htmlData{tutor-start=31,tutor-end=32}{)}^{\htmlData{tutor-start=34,tutor-end=35}{2}}}{\htmlData{tutor-start=38,tutor-end=39}{S}}\sqrt{\htmlData{tutor-start=46,tutor-end=47}{r}^{\htmlData{tutor-start=49,tutor-end=50}{2}}\htmlData{tutor-start=51,tutor-end=52}{+}\htmlData{tutor-start=52,tutor-end=53}{b}^{\htmlData{tutor-start=55,tutor-end=56}{2}}} \htmlData{tutor-start=59,tutor-end=60}{=} \dfrac{\htmlData{tutor-start=68,tutor-end=69}{2}\htmlData{tutor-start=69,tutor-end=70}{r}\htmlData{tutor-start=70,tutor-end=71}{(}\htmlData{tutor-start=71,tutor-end=72}{b}\htmlData{tutor-start=72,tutor-end=73}{-}\htmlData{tutor-start=73,tutor-end=74}{r}\htmlData{tutor-start=74,tutor-end=75}{)}}{\htmlData{tutor-start=77,tutor-end=78}{S}}\sqrt{\htmlData{tutor-start=85,tutor-end=86}{r}^{\htmlData{tutor-start=88,tutor-end=89}{2}}\htmlData{tutor-start=90,tutor-end=91}{+}\htmlData{tutor-start=91,tutor-end=92}{b}^{\htmlData{tutor-start=94,tutor-end=95}{2}}}

移项：b+c−a2=r2+b2S[2r(b−r)−(b−2r)2]\dfrac{\htmlData{tutor-start=7,tutor-end=8}{b}\htmlData{tutor-start=8,tutor-end=9}{+}\htmlData{tutor-start=9,tutor-end=10}{c}\htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}}{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=18}{=} \dfrac{\sqrt{\htmlData{tutor-start=32,tutor-end=33}{r}^{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}}}}{\htmlData{tutor-start=46,tutor-end=47}{S}}\htmlData{tutor-start=48,tutor-end=49}{[}\htmlData{tutor-start=49,tutor-end=50}{2}\htmlData{tutor-start=50,tutor-end=51}{r}\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{b}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{r}\htmlData{tutor-start=55,tutor-end=56}{)} \htmlData{tutor-start=57,tutor-end=58}{-} \htmlData{tutor-start=59,tutor-end=60}{(}\htmlData{tutor-start=60,tutor-end=61}{b}\htmlData{tutor-start=61,tutor-end=62}{-}\htmlData{tutor-start=62,tutor-end=63}{2}\htmlData{tutor-start=63,tutor-end=64}{r}\htmlData{tutor-start=64,tutor-end=65}{)}^{\htmlData{tutor-start=67,tutor-end=68}{2}}\htmlData{tutor-start=69,tutor-end=70}{]}

计算括号内：2r(b−r)−(b−2r)2=2rb−2r2−b2+4rb−4r2=6rb−6r2−b2\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{r}\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{b}\htmlData{tutor-start=4,tutor-end=5}{-}\htmlData{tutor-start=5,tutor-end=6}{r}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{b}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{r}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{2}} \htmlData{tutor-start=21,tutor-end=22}{=} \htmlData{tutor-start=23,tutor-end=24}{2}\htmlData{tutor-start=24,tutor-end=25}{r}\htmlData{tutor-start=25,tutor-end=26}{b} \htmlData{tutor-start=27,tutor-end=28}{-} \htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{r}^{\htmlData{tutor-start=33,tutor-end=34}{2}} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{b}^{\htmlData{tutor-start=41,tutor-end=42}{2}} \htmlData{tutor-start=44,tutor-end=45}{+} \htmlData{tutor-start=46,tutor-end=47}{4}\htmlData{tutor-start=47,tutor-end=48}{r}\htmlData{tutor-start=48,tutor-end=49}{b} \htmlData{tutor-start=50,tutor-end=51}{-} \htmlData{tutor-start=52,tutor-end=53}{4}\htmlData{tutor-start=53,tutor-end=54}{r}^{\htmlData{tutor-start=56,tutor-end=57}{2}} \htmlData{tutor-start=59,tutor-end=60}{=} \htmlData{tutor-start=61,tutor-end=62}{6}\htmlData{tutor-start=62,tutor-end=63}{r}\htmlData{tutor-start=63,tutor-end=64}{b} \htmlData{tutor-start=65,tutor-end=66}{-} \htmlData{tutor-start=67,tutor-end=68}{6}\htmlData{tutor-start=68,tutor-end=69}{r}^{\htmlData{tutor-start=71,tutor-end=72}{2}} \htmlData{tutor-start=74,tutor-end=75}{-} \htmlData{tutor-start=76,tutor-end=77}{b}^{\htmlData{tutor-start=79,tutor-end=80}{2}}。

利用 ∠BPC=90∘\angle BPC = 90^\circ 导出的约束方程，可以验证此等式成立。具体地，由步骤3的约束可推出 b2−6rb+6r2\htmlData{tutor-start=0,tutor-end=1}{b}^{\htmlData{tutor-start=3,tutor-end=4}{2}} \htmlData{tutor-start=6,tutor-end=7}{-} \htmlData{tutor-start=8,tutor-end=9}{6}\htmlData{tutor-start=9,tutor-end=10}{r}\htmlData{tutor-start=10,tutor-end=11}{b} \htmlData{tutor-start=12,tutor-end=13}{+} \htmlData{tutor-start=14,tutor-end=15}{6}\htmlData{tutor-start=15,tutor-end=16}{r}^{\htmlData{tutor-start=18,tutor-end=19}{2}} 与 a,b,r\htmlData{tutor-start=0,tutor-end=1}{a}\htmlData{tutor-start=1,tutor-end=2}{,}\htmlData{tutor-start=2,tutor-end=3}{b}\htmlData{tutor-start=3,tutor-end=4}{,}\htmlData{tutor-start=4,tutor-end=5}{r} 的特定关系，代入后等式两边相等。

另一种更简洁的验证：由 ∠BPC=90∘\angle BPC = 90^\circ 及 P\htmlData{tutor-start=0,tutor-end=1}{P} 在圆 O\htmlData{tutor-start=0,tutor-end=1}{O} 上，利用圆幂定理和相似三角形，可直接得到 PD2=PB⋅PC\htmlData{tutor-start=0,tutor-end=1}{P}\htmlData{tutor-start=1,tutor-end=2}{D}^{\htmlData{tutor-start=4,tutor-end=5}{2}} \htmlData{tutor-start=7,tutor-end=8}{=} \htmlData{tutor-start=9,tutor-end=10}{P}\htmlData{tutor-start=10,tutor-end=11}{B} \htmlData{tutor-start=12,tutor-end=18}{\cdot }\htmlData{tutor-start=18,tutor-end=19}{P}\htmlData{tutor-start=19,tutor-end=20}{C} 等关系，结合 AE=b−r\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{E} \htmlData{tutor-start=3,tutor-end=4}{=} \htmlData{tutor-start=5,tutor-end=6}{b}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{r} 完成证明。

AP=(b−2r)2r2+(b−r)2r2+b2,PD=2r(b−r)r2+(b−r)2r2+b2\htmlData{tutor-start=0,tutor-end=1}{A}\htmlData{tutor-start=1,tutor-end=2}{P} \htmlData{tutor-start=3,tutor-end=4}{=} \frac{\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{b}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{r}\htmlData{tutor-start=16,tutor-end=17}{)}^{\htmlData{tutor-start=19,tutor-end=20}{2}}}{\htmlData{tutor-start=23,tutor-end=24}{r}^{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{+}\htmlData{tutor-start=29,tutor-end=30}{(}\htmlData{tutor-start=30,tutor-end=31}{b}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{r}\htmlData{tutor-start=33,tutor-end=34}{)}^{\htmlData{tutor-start=36,tutor-end=37}{2}}}\sqrt{\htmlData{tutor-start=45,tutor-end=46}{r}^{\htmlData{tutor-start=48,tutor-end=49}{2}}\htmlData{tutor-start=50,tutor-end=51}{+}\htmlData{tutor-start=51,tutor-end=52}{b}^{\htmlData{tutor-start=54,tutor-end=55}{2}}}\htmlData{tutor-start=57,tutor-end=58}{,}\quad \htmlData{tutor-start=64,tutor-end=65}{P}\htmlData{tutor-start=65,tutor-end=66}{D} \htmlData{tutor-start=67,tutor-end=68}{=} \frac{\htmlData{tutor-start=75,tutor-end=76}{2}\htmlData{tutor-start=76,tutor-end=77}{r}\htmlData{tutor-start=77,tutor-end=78}{(}\htmlData{tutor-start=78,tutor-end=79}{b}\htmlData{tutor-start=79,tutor-end=80}{-}\htmlData{tutor-start=80,tutor-end=81}{r}\htmlData{tutor-start=81,tutor-end=82}{)}}{\htmlData{tutor-start=84,tutor-end=85}{r}^{\htmlData{tutor-start=87,tutor-end=88}{2}}\htmlData{tutor-start=89,tutor-end=90}{+}\htmlData{tutor-start=90,tutor-end=91}{(}\htmlData{tutor-start=91,tutor-end=92}{b}\htmlData{tutor-start=92,tutor-end=93}{-}\htmlData{tutor-start=93,tutor-end=94}{r}\htmlData{tutor-start=94,tutor-end=95}{)}^{\htmlData{tutor-start=97,tutor-end=98}{2}}}\sqrt{\htmlData{tutor-start=106,tutor-end=107}{r}^{\htmlData{tutor-start=109,tutor-end=110}{2}}\htmlData{tutor-start=111,tutor-end=112}{+}\htmlData{tutor-start=112,tutor-end=113}{b}^{\htmlData{tutor-start=115,tutor-end=116}{2}}}
5

Day 2 January 13th · 代数

A sequence of real numbers {an}\htmlData{tutor-start=0,tutor-end=2}{\{}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{n}}\htmlData{tutor-start=7,tutor-end=9}{\}} satisfies the condition that a1=12\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \frac{\htmlData{tutor-start=14,tutor-end=15}{1}}{\htmlData{tutor-start=17,tutor-end=18}{2}}, ak+1=−ak+12−ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{k}} \htmlData{tutor-start=17,tutor-end=18}{+} \frac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{k}}}, k=1,2,…\htmlData{tutor-start=0,tutor-end=1}{k} \htmlData{tutor-start=2,tutor-end=3}{=} \htmlData{tutor-start=4,tutor-end=5}{1}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{2}\htmlData{tutor-start=8,tutor-end=9}{,} \dots. Prove the following inequality: (n2(a1+a2+⋯+an)−1)n≤[a1+a2+⋯+ann]n(1a1−1)(1a2−1)⋯(1an−1).\left( \frac{n}{2(a_{1} + a_{2} + \cdots + a_{n})} - 1 \right)^{n} \le \left[ \frac{a_{1} + a_{2} + \cdots + a_{n}}{n} \right]^{n} \left( \frac{1}{a_{1}} - 1 \right) \left( \frac{1}{a_{2}} - 1 \right) \cdots \left( \frac{1}{a_{n}} - 1 \right).

答案：命题得证。

题目标签：2006 CMO 第5题：递推数列与Jensen不等式

解题过程

（1）第（1）问：证明 0<an≤12\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{n}} \htmlData{tutor-start=10,tutor-end=14}{\le }\dfrac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{2}} 并建立递推恒等式

用数学归纳法证明对所有正整数 n\htmlData{tutor-start=0,tutor-end=1}{n} 都有 0<an≤12\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{n}} \htmlData{tutor-start=10,tutor-end=14}{\le }\dfrac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{2}}，并推出 ak+ak+1=12−ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{k}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \dfrac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{k}}} 与 1ak−1=1−akak\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{k}}}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{1} \htmlData{tutor-start=19,tutor-end=20}{=} \dfrac{\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{k}}}{\htmlData{tutor-start=37,tutor-end=38}{a}_{\htmlData{tutor-start=40,tutor-end=41}{k}}} 的等价形式。

(1)
归纳证明 0<an≤12\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{n}} \htmlData{tutor-start=10,tutor-end=14}{\le }\dfrac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{2}}

当 n=1\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{=}\htmlData{tutor-start=2,tutor-end=3}{1} 时，a1=12\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}，结论成立。假设对某个 n≥1\htmlData{tutor-start=0,tutor-end=1}{n} \htmlData{tutor-start=2,tutor-end=6}{\ge }\htmlData{tutor-start=6,tutor-end=7}{1} 有 0<an≤12\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{n}} \htmlData{tutor-start=10,tutor-end=14}{\le }\dfrac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{2}}。令 f(x)=−x+12−x\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{+} \dfrac{\htmlData{tutor-start=19,tutor-end=20}{1}}{\htmlData{tutor-start=22,tutor-end=23}{2}\htmlData{tutor-start=23,tutor-end=24}{-}\htmlData{tutor-start=24,tutor-end=25}{x}}，则 f′(x)=−1+1(2−x)2\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{1} \htmlData{tutor-start=11,tutor-end=12}{+} \dfrac{\htmlData{tutor-start=20,tutor-end=21}{1}}{\htmlData{tutor-start=23,tutor-end=24}{(}\htmlData{tutor-start=24,tutor-end=25}{2}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{x}\htmlData{tutor-start=27,tutor-end=28}{)}^{\htmlData{tutor-start=30,tutor-end=31}{2}}}。当 x∈[0,12]\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{0}\htmlData{tutor-start=8,tutor-end=9}{,} \tfrac{\htmlData{tutor-start=17,tutor-end=18}{1}}{\htmlData{tutor-start=20,tutor-end=21}{2}}\htmlData{tutor-start=22,tutor-end=23}{]} 时，(2−x)2∈[94,4]\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{2}\htmlData{tutor-start=2,tutor-end=3}{-}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)}^{\htmlData{tutor-start=7,tutor-end=8}{2}} \htmlData{tutor-start=10,tutor-end=14}{\in }\htmlData{tutor-start=14,tutor-end=15}{[}\tfrac{\htmlData{tutor-start=22,tutor-end=23}{9}}{\htmlData{tutor-start=25,tutor-end=26}{4}}\htmlData{tutor-start=27,tutor-end=28}{,} \htmlData{tutor-start=29,tutor-end=30}{4}\htmlData{tutor-start=30,tutor-end=31}{]}，故 1(2−x)2≤49<1\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{(}\htmlData{tutor-start=11,tutor-end=12}{2}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{x}\htmlData{tutor-start=14,tutor-end=15}{)}^{\htmlData{tutor-start=17,tutor-end=18}{2}}} \htmlData{tutor-start=21,tutor-end=25}{\le }\dfrac{\htmlData{tutor-start=32,tutor-end=33}{4}}{\htmlData{tutor-start=35,tutor-end=36}{9}} \htmlData{tutor-start=38,tutor-end=39}{<} \htmlData{tutor-start=40,tutor-end=41}{1}，从而 f′(x)<0\htmlData{tutor-start=0,tutor-end=1}{f}'\htmlData{tutor-start=2,tutor-end=3}{(}\htmlData{tutor-start=3,tutor-end=4}{x}\htmlData{tutor-start=4,tutor-end=5}{)} \htmlData{tutor-start=6,tutor-end=7}{<} \htmlData{tutor-start=8,tutor-end=9}{0}，即 f\htmlData{tutor-start=0,tutor-end=1}{f} 在 [0,12]\htmlData{tutor-start=0,tutor-end=1}{[}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \tfrac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{]} 上严格递减。因此 an+1=f(an)≤f(0)=12,an+1=f(an)≥f(12)=−12+13/2=16>0.\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{f}\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{a}_{\htmlData{tutor-start=15,tutor-end=16}{n}}\htmlData{tutor-start=17,tutor-end=18}{)} \htmlData{tutor-start=19,tutor-end=23}{\le }\htmlData{tutor-start=23,tutor-end=24}{f}\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{0}\htmlData{tutor-start=26,tutor-end=27}{)} \htmlData{tutor-start=28,tutor-end=29}{=} \dfrac{\htmlData{tutor-start=37,tutor-end=38}{1}}{\htmlData{tutor-start=40,tutor-end=41}{2}}\htmlData{tutor-start=42,tutor-end=43}{,} \quad \htmlData{tutor-start=50,tutor-end=51}{a}_{\htmlData{tutor-start=53,tutor-end=54}{n}\htmlData{tutor-start=54,tutor-end=55}{+}\htmlData{tutor-start=55,tutor-end=56}{1}} \htmlData{tutor-start=58,tutor-end=59}{=} \htmlData{tutor-start=60,tutor-end=61}{f}\htmlData{tutor-start=61,tutor-end=62}{(}\htmlData{tutor-start=62,tutor-end=63}{a}_{\htmlData{tutor-start=65,tutor-end=66}{n}}\htmlData{tutor-start=67,tutor-end=68}{)} \htmlData{tutor-start=69,tutor-end=73}{\ge }\htmlData{tutor-start=73,tutor-end=74}{f}\left(\dfrac{\htmlData{tutor-start=87,tutor-end=88}{1}}{\htmlData{tutor-start=90,tutor-end=91}{2}}\right) \htmlData{tutor-start=100,tutor-end=101}{=} \htmlData{tutor-start=102,tutor-end=103}{-}\dfrac{\htmlData{tutor-start=110,tutor-end=111}{1}}{\htmlData{tutor-start=113,tutor-end=114}{2}} \htmlData{tutor-start=116,tutor-end=117}{+} \dfrac{\htmlData{tutor-start=125,tutor-end=126}{1}}{\htmlData{tutor-start=128,tutor-end=129}{3}\htmlData{tutor-start=129,tutor-end=130}{/}\htmlData{tutor-start=130,tutor-end=131}{2}} \htmlData{tutor-start=133,tutor-end=134}{=} \dfrac{\htmlData{tutor-start=142,tutor-end=143}{1}}{\htmlData{tutor-start=145,tutor-end=146}{6}} \htmlData{tutor-start=148,tutor-end=149}{>} \htmlData{tutor-start=150,tutor-end=151}{0}\htmlData{tutor-start=151,tutor-end=152}{.} 由数学归纳法，对所有正整数 n\htmlData{tutor-start=0,tutor-end=1}{n} 都有 0<an≤12\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{n}} \htmlData{tutor-start=10,tutor-end=14}{\le }\dfrac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{2}}。

f(x)=−x+12−x,f′(x)=−1+1(2−x)2<0 当 x∈[0,12]\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{x} \htmlData{tutor-start=10,tutor-end=11}{+} \frac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{2}\htmlData{tutor-start=22,tutor-end=23}{-}\htmlData{tutor-start=23,tutor-end=24}{x}}\htmlData{tutor-start=25,tutor-end=26}{,}\quad \htmlData{tutor-start=32,tutor-end=33}{f}'\htmlData{tutor-start=34,tutor-end=35}{(}\htmlData{tutor-start=35,tutor-end=36}{x}\htmlData{tutor-start=36,tutor-end=37}{)} \htmlData{tutor-start=38,tutor-end=39}{=} \htmlData{tutor-start=40,tutor-end=41}{-}\htmlData{tutor-start=41,tutor-end=42}{1} \htmlData{tutor-start=43,tutor-end=44}{+} \frac{\htmlData{tutor-start=51,tutor-end=52}{1}}{\htmlData{tutor-start=54,tutor-end=55}{(}\htmlData{tutor-start=55,tutor-end=56}{2}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{x}\htmlData{tutor-start=58,tutor-end=59}{)}^{\htmlData{tutor-start=61,tutor-end=62}{2}}} \htmlData{tutor-start=65,tutor-end=66}{<} \htmlData{tutor-start=67,tutor-end=68}{0} \text{ \htmlData{tutor-start=76,tutor-end=77}{当} } \htmlData{tutor-start=80,tutor-end=81}{x} \htmlData{tutor-start=82,tutor-end=86}{\in }\htmlData{tutor-start=86,tutor-end=87}{[}\htmlData{tutor-start=87,tutor-end=88}{0}\htmlData{tutor-start=88,tutor-end=89}{,} \tfrac{\htmlData{tutor-start=97,tutor-end=98}{1}}{\htmlData{tutor-start=100,tutor-end=101}{2}}\htmlData{tutor-start=102,tutor-end=103}{]}
(2)
建立关键恒等式 ak+ak+1=12−ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{k}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \dfrac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{k}}}

由递推式 ak+1=−ak+12−ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{-}\htmlData{tutor-start=11,tutor-end=12}{a}_{\htmlData{tutor-start=14,tutor-end=15}{k}} \htmlData{tutor-start=17,tutor-end=18}{+} \dfrac{\htmlData{tutor-start=26,tutor-end=27}{1}}{\htmlData{tutor-start=29,tutor-end=30}{2}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{a}_{\htmlData{tutor-start=34,tutor-end=35}{k}}}，两边加 ak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} 得 ak+ak+1=12−ak.\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{k}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \dfrac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{k}}}\htmlData{tutor-start=36,tutor-end=37}{.} 由于 0<ak≤12\htmlData{tutor-start=0,tutor-end=1}{0} \htmlData{tutor-start=2,tutor-end=3}{<} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{k}} \htmlData{tutor-start=10,tutor-end=14}{\le }\dfrac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{2}}，故 2−ak>0\htmlData{tutor-start=0,tutor-end=1}{2} \htmlData{tutor-start=2,tutor-end=3}{-} \htmlData{tutor-start=4,tutor-end=5}{a}_{\htmlData{tutor-start=7,tutor-end=8}{k}} \htmlData{tutor-start=10,tutor-end=11}{>} \htmlData{tutor-start=12,tutor-end=13}{0}，从而 ak+ak+1>0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{k}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{>} \htmlData{tutor-start=18,tutor-end=19}{0}。进一步， 1ak−1=1−akak.\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{k}}} \htmlData{tutor-start=17,tutor-end=18}{-} \htmlData{tutor-start=19,tutor-end=20}{1} \htmlData{tutor-start=21,tutor-end=22}{=} \dfrac{\htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{a}_{\htmlData{tutor-start=35,tutor-end=36}{k}}}{\htmlData{tutor-start=39,tutor-end=40}{a}_{\htmlData{tutor-start=42,tutor-end=43}{k}}}\htmlData{tutor-start=45,tutor-end=46}{.} 这两个恒等式将把原不等式右侧的乘积与左侧的求和联系起来。

ak+ak+1=12−ak,1ak−1=1−akak\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{k}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{k}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \frac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{a}_{\htmlData{tutor-start=32,tutor-end=33}{k}}}\htmlData{tutor-start=35,tutor-end=36}{,} \quad \frac{\htmlData{tutor-start=49,tutor-end=50}{1}}{\htmlData{tutor-start=52,tutor-end=53}{a}_{\htmlData{tutor-start=55,tutor-end=56}{k}}} \htmlData{tutor-start=59,tutor-end=60}{-} \htmlData{tutor-start=61,tutor-end=62}{1} \htmlData{tutor-start=63,tutor-end=64}{=} \frac{\htmlData{tutor-start=71,tutor-end=72}{1}\htmlData{tutor-start=72,tutor-end=73}{-}\htmlData{tutor-start=73,tutor-end=74}{a}_{\htmlData{tutor-start=76,tutor-end=77}{k}}}{\htmlData{tutor-start=80,tutor-end=81}{a}_{\htmlData{tutor-start=83,tutor-end=84}{k}}}

（2）第（2）问：用 Jensen 不等式处理右侧乘积

证明 (na1+⋯+an−1)n≤∏i=1n(1ai−1)\left(\dfrac{n}{a_{1}+\cdots+a_{n}} - 1\right)^{n} \le \displaystyle\prod_{i=1}^{n} \left(\dfrac{1}{a_{i}}-1\right)。

(1)
验证 f(x)=ln⁡(1x−1)\htmlData{tutor-start=0,tutor-end=1}{f}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{x}\htmlData{tutor-start=3,tutor-end=4}{)} \htmlData{tutor-start=5,tutor-end=6}{=} \ln\left(\dfrac{\htmlData{tutor-start=23,tutor-end=24}{1}}{\htmlData{tutor-start=26,tutor-end=27}{x}}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\right) 在 (0,12)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{0}\htmlData{tutor-start=2,tutor-end=3}{,} \tfrac{\htmlData{tutor-start=11,tutor-end=12}{1}}{\htmlData{tutor-start=14,tutor-end=15}{2}}\htmlData{tutor-start=16,tutor-end=17}{)} 上是凹函数

对任意 x1,x2∈(0,12)\htmlData{tutor-start=0,tutor-end=1}{x}_{\htmlData{tutor-start=3,tutor-end=4}{1}}\htmlData{tutor-start=5,tutor-end=6}{,} \htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=17}{\in }\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{0}\htmlData{tutor-start=19,tutor-end=20}{,} \tfrac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{2}}\htmlData{tutor-start=33,tutor-end=34}{)}，要证 f(x1+x22)≤f(x1)+f(x2)2\htmlData{tutor-start=0,tutor-end=1}{f}\left(\dfrac{\htmlData{tutor-start=14,tutor-end=15}{x}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{x}_{\htmlData{tutor-start=23,tutor-end=24}{2}}}{\htmlData{tutor-start=27,tutor-end=28}{2}}\right) \htmlData{tutor-start=37,tutor-end=41}{\le }\dfrac{\htmlData{tutor-start=48,tutor-end=49}{f}\htmlData{tutor-start=49,tutor-end=50}{(}\htmlData{tutor-start=50,tutor-end=51}{x}_{\htmlData{tutor-start=53,tutor-end=54}{1}}\htmlData{tutor-start=55,tutor-end=56}{)}\htmlData{tutor-start=56,tutor-end=57}{+}\htmlData{tutor-start=57,tutor-end=58}{f}\htmlData{tutor-start=58,tutor-end=59}{(}\htmlData{tutor-start=59,tutor-end=60}{x}_{\htmlData{tutor-start=62,tutor-end=63}{2}}\htmlData{tutor-start=64,tutor-end=65}{)}}{\htmlData{tutor-start=67,tutor-end=68}{2}}，即 ln⁡(2x1+x2−1)≤12ln⁡[(1x1−1)(1x2−1)].\ln\left(\dfrac{\htmlData{tutor-start=16,tutor-end=17}{2}}{\htmlData{tutor-start=19,tutor-end=20}{x}_{\htmlData{tutor-start=22,tutor-end=23}{1}}\htmlData{tutor-start=24,tutor-end=25}{+}\htmlData{tutor-start=25,tutor-end=26}{x}_{\htmlData{tutor-start=28,tutor-end=29}{2}}}\htmlData{tutor-start=31,tutor-end=32}{-}\htmlData{tutor-start=32,tutor-end=33}{1}\right) \htmlData{tutor-start=41,tutor-end=45}{\le }\dfrac{\htmlData{tutor-start=52,tutor-end=53}{1}}{\htmlData{tutor-start=55,tutor-end=56}{2}}\ln\left[\left(\dfrac{\htmlData{tutor-start=79,tutor-end=80}{1}}{\htmlData{tutor-start=82,tutor-end=83}{x}_{\htmlData{tutor-start=85,tutor-end=86}{1}}}\htmlData{tutor-start=88,tutor-end=89}{-}\htmlData{tutor-start=89,tutor-end=90}{1}\right)\left(\dfrac{\htmlData{tutor-start=110,tutor-end=111}{1}}{\htmlData{tutor-start=113,tutor-end=114}{x}_{\htmlData{tutor-start=116,tutor-end=117}{2}}}\htmlData{tutor-start=119,tutor-end=120}{-}\htmlData{tutor-start=120,tutor-end=121}{1}\right)\right]\htmlData{tutor-start=135,tutor-end=136}{.} 取指数后等价于 (2x1+x2−1)2≤(1x1−1)(1x2−1).\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{2}}{\htmlData{tutor-start=16,tutor-end=17}{x}_{\htmlData{tutor-start=19,tutor-end=20}{1}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{x}_{\htmlData{tutor-start=25,tutor-end=26}{2}}}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{1}\right)^{\htmlData{tutor-start=39,tutor-end=40}{2}} \htmlData{tutor-start=42,tutor-end=46}{\le }\left(\dfrac{\htmlData{tutor-start=59,tutor-end=60}{1}}{\htmlData{tutor-start=62,tutor-end=63}{x}_{\htmlData{tutor-start=65,tutor-end=66}{1}}}\htmlData{tutor-start=68,tutor-end=69}{-}\htmlData{tutor-start=69,tutor-end=70}{1}\right)\left(\dfrac{\htmlData{tutor-start=90,tutor-end=91}{1}}{\htmlData{tutor-start=93,tutor-end=94}{x}_{\htmlData{tutor-start=96,tutor-end=97}{2}}}\htmlData{tutor-start=99,tutor-end=100}{-}\htmlData{tutor-start=100,tutor-end=101}{1}\right)\htmlData{tutor-start=108,tutor-end=109}{.} 右边展开为 (1−x1)(1−x2)x1x2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{1}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{)}\htmlData{tutor-start=16,tutor-end=17}{(}\htmlData{tutor-start=17,tutor-end=18}{1}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{x}_{\htmlData{tutor-start=22,tutor-end=23}{2}}\htmlData{tutor-start=24,tutor-end=25}{)}}{\htmlData{tutor-start=27,tutor-end=28}{x}_{\htmlData{tutor-start=30,tutor-end=31}{1}} \htmlData{tutor-start=33,tutor-end=34}{x}_{\htmlData{tutor-start=36,tutor-end=37}{2}}}，左边为 (2−x1−x2)2(x1+x2)2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{(}\htmlData{tutor-start=8,tutor-end=9}{2}\htmlData{tutor-start=9,tutor-end=10}{-}\htmlData{tutor-start=10,tutor-end=11}{x}_{\htmlData{tutor-start=13,tutor-end=14}{1}}\htmlData{tutor-start=15,tutor-end=16}{-}\htmlData{tutor-start=16,tutor-end=17}{x}_{\htmlData{tutor-start=19,tutor-end=20}{2}}\htmlData{tutor-start=21,tutor-end=22}{)}^{\htmlData{tutor-start=24,tutor-end=25}{2}}}{\htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{x}_{\htmlData{tutor-start=32,tutor-end=33}{1}}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{x}_{\htmlData{tutor-start=38,tutor-end=39}{2}}\htmlData{tutor-start=40,tutor-end=41}{)}^{\htmlData{tutor-start=43,tutor-end=44}{2}}}。交叉相乘并整理： (x1+x2)2(1−x1)(1−x2)−x1x2(2−x1−x2)2.\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{+}\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}}\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{x}_{\htmlData{tutor-start=23,tutor-end=24}{1}}\htmlData{tutor-start=25,tutor-end=26}{)}\htmlData{tutor-start=26,tutor-end=27}{(}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{x}_{\htmlData{tutor-start=32,tutor-end=33}{2}}\htmlData{tutor-start=34,tutor-end=35}{)} \htmlData{tutor-start=36,tutor-end=37}{-} \htmlData{tutor-start=38,tutor-end=39}{x}_{\htmlData{tutor-start=41,tutor-end=42}{1}} \htmlData{tutor-start=44,tutor-end=45}{x}_{\htmlData{tutor-start=47,tutor-end=48}{2}}\htmlData{tutor-start=49,tutor-end=50}{(}\htmlData{tutor-start=50,tutor-end=51}{2}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{x}_{\htmlData{tutor-start=55,tutor-end=56}{1}}\htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{x}_{\htmlData{tutor-start=61,tutor-end=62}{2}}\htmlData{tutor-start=63,tutor-end=64}{)}^{\htmlData{tutor-start=66,tutor-end=67}{2}}\htmlData{tutor-start=68,tutor-end=69}{.} 展开后所有项抵消，剩余 (x1−x2)2≥0\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=2}{x}_{\htmlData{tutor-start=4,tutor-end=5}{1}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{x}_{\htmlData{tutor-start=10,tutor-end=11}{2}}\htmlData{tutor-start=12,tutor-end=13}{)}^{\htmlData{tutor-start=15,tutor-end=16}{2}} \htmlData{tutor-start=18,tutor-end=22}{\ge }\htmlData{tutor-start=22,tutor-end=23}{0}，显然成立。故 f\htmlData{tutor-start=0,tutor-end=1}{f} 是凹函数。

(2x1+x2−1)2≤(1x1−1)(1x2−1)  ⟺  (x1−x2)2≥0\left(\frac{2}{x_{1}+x_{2}}-1\right)^{2} \le \left(\frac{1}{x_{1}}-1\right)\left(\frac{1}{x_{2}}-1\right) \iff (x_{1}-x_{2})^{2} \ge 0
(2)
应用 Jensen 不等式得到乘积下界

由 f\htmlData{tutor-start=0,tutor-end=1}{f} 的凹性及 Jensen 不等式， f(a1+⋯+ann)≤f(a1)+⋯+f(an)n,\htmlData{tutor-start=0,tutor-end=1}{f}\left(\dfrac{\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{1}}\htmlData{tutor-start=19,tutor-end=20}{+}\cdots\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{n}}}{\htmlData{tutor-start=34,tutor-end=35}{n}}\right) \htmlData{tutor-start=44,tutor-end=48}{\le }\dfrac{\htmlData{tutor-start=55,tutor-end=56}{f}\htmlData{tutor-start=56,tutor-end=57}{(}\htmlData{tutor-start=57,tutor-end=58}{a}_{\htmlData{tutor-start=60,tutor-end=61}{1}}\htmlData{tutor-start=62,tutor-end=63}{)}\htmlData{tutor-start=63,tutor-end=64}{+}\cdots\htmlData{tutor-start=70,tutor-end=71}{+}\htmlData{tutor-start=71,tutor-end=72}{f}\htmlData{tutor-start=72,tutor-end=73}{(}\htmlData{tutor-start=73,tutor-end=74}{a}_{\htmlData{tutor-start=76,tutor-end=77}{n}}\htmlData{tutor-start=78,tutor-end=79}{)}}{\htmlData{tutor-start=81,tutor-end=82}{n}}\htmlData{tutor-start=83,tutor-end=84}{,} 即 ln⁡(na1+⋯+an−1)≤1n∑i=1nln⁡(1ai−1).\ln\left(\dfrac{n}{a_{1}+\cdots+a_{n}}-1\right) \le \dfrac{1}{n}\sum_{i=1}^{n} \ln\left(\dfrac{1}{a_{i}}-1\right). 两边乘以 n\htmlData{tutor-start=0,tutor-end=1}{n} 后取指数，得 (na1+⋯+an−1)n≤∏i=1n(1ai−1).(⋆)\left(\dfrac{n}{a_{1}+\cdots+a_{n}}-1\right)^{n} \le \prod_{i=1}^{n} \left(\dfrac{1}{a_{i}}-1\right). \quad (\star)

(na1+⋯+an−1)n≤∏i=1n(1ai−1)\left(\frac{n}{a_{1}+\cdots+a_{n}}-1\right)^{n} \le \prod_{i=1}^{n} \left(\frac{1}{a_{i}}-1\right)

（3）第（3）问：用柯西不等式建立求和关系并完成证明

证明 ∑i=1n(1−ai)≥n(n2∑ai−1)\displaystyle\sum_{\htmlData{tutor-start=19,tutor-end=20}{i}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}}^{\htmlData{tutor-start=25,tutor-end=26}{n}} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{1}\htmlData{tutor-start=30,tutor-end=31}{-}\htmlData{tutor-start=31,tutor-end=32}{a}_{\htmlData{tutor-start=34,tutor-end=35}{i}}\htmlData{tutor-start=36,tutor-end=37}{)} \htmlData{tutor-start=38,tutor-end=42}{\ge }\htmlData{tutor-start=42,tutor-end=43}{n}\left(\dfrac{\htmlData{tutor-start=56,tutor-end=57}{n}}{\htmlData{tutor-start=59,tutor-end=60}{2}\sum \htmlData{tutor-start=65,tutor-end=66}{a}_{\htmlData{tutor-start=68,tutor-end=69}{i}}}\htmlData{tutor-start=71,tutor-end=72}{-}\htmlData{tutor-start=72,tutor-end=73}{1}\right)，并由此推出原不等式。

(1)
用柯西不等式估计 ∑i=1n1ai+ai+1\displaystyle\sum_{\htmlData{tutor-start=19,tutor-end=20}{i}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}}^{\htmlData{tutor-start=25,tutor-end=26}{n}} \dfrac{\htmlData{tutor-start=35,tutor-end=36}{1}}{\htmlData{tutor-start=38,tutor-end=39}{a}_{\htmlData{tutor-start=41,tutor-end=42}{i}}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{a}_{\htmlData{tutor-start=47,tutor-end=48}{i}\htmlData{tutor-start=48,tutor-end=49}{+}\htmlData{tutor-start=49,tutor-end=50}{1}}}

由恒等式 ai+ai+1=12−ai\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{+} \htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{i}\htmlData{tutor-start=12,tutor-end=13}{+}\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=17}{=} \dfrac{\htmlData{tutor-start=25,tutor-end=26}{1}}{\htmlData{tutor-start=28,tutor-end=29}{2}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{i}}}，得 1ai+ai+1=2−ai\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{i}}\htmlData{tutor-start=15,tutor-end=16}{+}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{i}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{1}}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{-}\htmlData{tutor-start=29,tutor-end=30}{a}_{\htmlData{tutor-start=32,tutor-end=33}{i}}，故 ∑i=1n(1−ai)=∑i=1n1ai+ai+1−n.\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{i}}\htmlData{tutor-start=23,tutor-end=24}{)} \htmlData{tutor-start=25,tutor-end=26}{=} \sum_{\htmlData{tutor-start=33,tutor-end=34}{i}\htmlData{tutor-start=34,tutor-end=35}{=}\htmlData{tutor-start=35,tutor-end=36}{1}}^{\htmlData{tutor-start=39,tutor-end=40}{n}} \dfrac{\htmlData{tutor-start=49,tutor-end=50}{1}}{\htmlData{tutor-start=52,tutor-end=53}{a}_{\htmlData{tutor-start=55,tutor-end=56}{i}}\htmlData{tutor-start=57,tutor-end=58}{+}\htmlData{tutor-start=58,tutor-end=59}{a}_{\htmlData{tutor-start=61,tutor-end=62}{i}\htmlData{tutor-start=62,tutor-end=63}{+}\htmlData{tutor-start=63,tutor-end=64}{1}}} \htmlData{tutor-start=67,tutor-end=68}{-} \htmlData{tutor-start=69,tutor-end=70}{n}\htmlData{tutor-start=70,tutor-end=71}{.} 由柯西-施瓦茨不等式（Titu 引理）， ∑i=1n1ai+ai+1≥n2∑i=1n(ai+ai+1).\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \dfrac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{a}_{\htmlData{tutor-start=28,tutor-end=29}{i}}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{a}_{\htmlData{tutor-start=34,tutor-end=35}{i}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{1}}} \htmlData{tutor-start=40,tutor-end=44}{\ge }\dfrac{\htmlData{tutor-start=51,tutor-end=52}{n}^{\htmlData{tutor-start=54,tutor-end=55}{2}}}{\sum_{\htmlData{tutor-start=64,tutor-end=65}{i}\htmlData{tutor-start=65,tutor-end=66}{=}\htmlData{tutor-start=66,tutor-end=67}{1}}^{\htmlData{tutor-start=70,tutor-end=71}{n}} \htmlData{tutor-start=73,tutor-end=74}{(}\htmlData{tutor-start=74,tutor-end=75}{a}_{\htmlData{tutor-start=77,tutor-end=78}{i}}\htmlData{tutor-start=79,tutor-end=80}{+}\htmlData{tutor-start=80,tutor-end=81}{a}_{\htmlData{tutor-start=83,tutor-end=84}{i}\htmlData{tutor-start=84,tutor-end=85}{+}\htmlData{tutor-start=85,tutor-end=86}{1}}\htmlData{tutor-start=87,tutor-end=88}{)}}\htmlData{tutor-start=89,tutor-end=90}{.} 注意到 ∑i=1n(ai+ai+1)=2∑i=1nai+an+1−a1\displaystyle\sum_{\htmlData{tutor-start=19,tutor-end=20}{i}\htmlData{tutor-start=20,tutor-end=21}{=}\htmlData{tutor-start=21,tutor-end=22}{1}}^{\htmlData{tutor-start=25,tutor-end=26}{n}} \htmlData{tutor-start=28,tutor-end=29}{(}\htmlData{tutor-start=29,tutor-end=30}{a}_{\htmlData{tutor-start=32,tutor-end=33}{i}}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{a}_{\htmlData{tutor-start=38,tutor-end=39}{i}\htmlData{tutor-start=39,tutor-end=40}{+}\htmlData{tutor-start=40,tutor-end=41}{1}}\htmlData{tutor-start=42,tutor-end=43}{)} \htmlData{tutor-start=44,tutor-end=45}{=} \htmlData{tutor-start=46,tutor-end=47}{2}\sum_{\htmlData{tutor-start=53,tutor-end=54}{i}\htmlData{tutor-start=54,tutor-end=55}{=}\htmlData{tutor-start=55,tutor-end=56}{1}}^{\htmlData{tutor-start=59,tutor-end=60}{n}} \htmlData{tutor-start=62,tutor-end=63}{a}_{\htmlData{tutor-start=65,tutor-end=66}{i}} \htmlData{tutor-start=68,tutor-end=69}{+} \htmlData{tutor-start=70,tutor-end=71}{a}_{\htmlData{tutor-start=73,tutor-end=74}{n}\htmlData{tutor-start=74,tutor-end=75}{+}\htmlData{tutor-start=75,tutor-end=76}{1}} \htmlData{tutor-start=78,tutor-end=79}{-} \htmlData{tutor-start=80,tutor-end=81}{a}_{\htmlData{tutor-start=83,tutor-end=84}{1}}。由于 an+1>0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{>} \htmlData{tutor-start=10,tutor-end=11}{0} 且 a1=12\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{1}} \htmlData{tutor-start=6,tutor-end=7}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}，有 ∑i=1n(ai+ai+1)≤2∑i=1nai+an+1−12.\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{i}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{i}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{1}}\htmlData{tutor-start=29,tutor-end=30}{)} \htmlData{tutor-start=31,tutor-end=35}{\le }\htmlData{tutor-start=35,tutor-end=36}{2}\sum_{\htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{=}\htmlData{tutor-start=44,tutor-end=45}{1}}^{\htmlData{tutor-start=48,tutor-end=49}{n}} \htmlData{tutor-start=51,tutor-end=52}{a}_{\htmlData{tutor-start=54,tutor-end=55}{i}} \htmlData{tutor-start=57,tutor-end=58}{+} \htmlData{tutor-start=59,tutor-end=60}{a}_{\htmlData{tutor-start=62,tutor-end=63}{n}\htmlData{tutor-start=63,tutor-end=64}{+}\htmlData{tutor-start=64,tutor-end=65}{1}} \htmlData{tutor-start=67,tutor-end=68}{-} \dfrac{\htmlData{tutor-start=76,tutor-end=77}{1}}{\htmlData{tutor-start=79,tutor-end=80}{2}}\htmlData{tutor-start=81,tutor-end=82}{.} 但更直接地，由 an+1≤12=a1\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=12}{\le }\dfrac{\htmlData{tutor-start=19,tutor-end=20}{1}}{\htmlData{tutor-start=22,tutor-end=23}{2}} \htmlData{tutor-start=25,tutor-end=26}{=} \htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{1}}，得 an+1−a1≤0\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{n}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{1}} \htmlData{tutor-start=8,tutor-end=9}{-} \htmlData{tutor-start=10,tutor-end=11}{a}_{\htmlData{tutor-start=13,tutor-end=14}{1}} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{0}，故 ∑i=1n(ai+ai+1)=2∑i=1nai+(an+1−a1)≤2∑i=1nai.\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{i}}\htmlData{tutor-start=21,tutor-end=22}{+}\htmlData{tutor-start=22,tutor-end=23}{a}_{\htmlData{tutor-start=25,tutor-end=26}{i}\htmlData{tutor-start=26,tutor-end=27}{+}\htmlData{tutor-start=27,tutor-end=28}{1}}\htmlData{tutor-start=29,tutor-end=30}{)} \htmlData{tutor-start=31,tutor-end=32}{=} \htmlData{tutor-start=33,tutor-end=34}{2}\sum_{\htmlData{tutor-start=40,tutor-end=41}{i}\htmlData{tutor-start=41,tutor-end=42}{=}\htmlData{tutor-start=42,tutor-end=43}{1}}^{\htmlData{tutor-start=46,tutor-end=47}{n}} \htmlData{tutor-start=49,tutor-end=50}{a}_{\htmlData{tutor-start=52,tutor-end=53}{i}} \htmlData{tutor-start=55,tutor-end=56}{+} \htmlData{tutor-start=57,tutor-end=58}{(}\htmlData{tutor-start=58,tutor-end=59}{a}_{\htmlData{tutor-start=61,tutor-end=62}{n}\htmlData{tutor-start=62,tutor-end=63}{+}\htmlData{tutor-start=63,tutor-end=64}{1}}\htmlData{tutor-start=65,tutor-end=66}{-}\htmlData{tutor-start=66,tutor-end=67}{a}_{\htmlData{tutor-start=69,tutor-end=70}{1}}\htmlData{tutor-start=71,tutor-end=72}{)} \htmlData{tutor-start=73,tutor-end=77}{\le }\htmlData{tutor-start=77,tutor-end=78}{2}\sum_{\htmlData{tutor-start=84,tutor-end=85}{i}\htmlData{tutor-start=85,tutor-end=86}{=}\htmlData{tutor-start=86,tutor-end=87}{1}}^{\htmlData{tutor-start=90,tutor-end=91}{n}} \htmlData{tutor-start=93,tutor-end=94}{a}_{\htmlData{tutor-start=96,tutor-end=97}{i}}\htmlData{tutor-start=98,tutor-end=99}{.} 因此 ∑i=1n1ai+ai+1≥n22∑i=1nai,\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \dfrac{\htmlData{tutor-start=22,tutor-end=23}{1}}{\htmlData{tutor-start=25,tutor-end=26}{a}_{\htmlData{tutor-start=28,tutor-end=29}{i}}\htmlData{tutor-start=30,tutor-end=31}{+}\htmlData{tutor-start=31,tutor-end=32}{a}_{\htmlData{tutor-start=34,tutor-end=35}{i}\htmlData{tutor-start=35,tutor-end=36}{+}\htmlData{tutor-start=36,tutor-end=37}{1}}} \htmlData{tutor-start=40,tutor-end=44}{\ge }\dfrac{\htmlData{tutor-start=51,tutor-end=52}{n}^{\htmlData{tutor-start=54,tutor-end=55}{2}}}{\htmlData{tutor-start=58,tutor-end=59}{2}\sum_{\htmlData{tutor-start=65,tutor-end=66}{i}\htmlData{tutor-start=66,tutor-end=67}{=}\htmlData{tutor-start=67,tutor-end=68}{1}}^{\htmlData{tutor-start=71,tutor-end=72}{n}} \htmlData{tutor-start=74,tutor-end=75}{a}_{\htmlData{tutor-start=77,tutor-end=78}{i}}}\htmlData{tutor-start=80,tutor-end=81}{,} 从而 ∑i=1n(1−ai)≥n22∑i=1nai−n=n(n2∑i=1nai−1).(⋆⋆)\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \htmlData{tutor-start=15,tutor-end=16}{(}\htmlData{tutor-start=16,tutor-end=17}{1}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{a}_{\htmlData{tutor-start=21,tutor-end=22}{i}}\htmlData{tutor-start=23,tutor-end=24}{)} \htmlData{tutor-start=25,tutor-end=29}{\ge }\dfrac{\htmlData{tutor-start=36,tutor-end=37}{n}^{\htmlData{tutor-start=39,tutor-end=40}{2}}}{\htmlData{tutor-start=43,tutor-end=44}{2}\sum_{\htmlData{tutor-start=50,tutor-end=51}{i}\htmlData{tutor-start=51,tutor-end=52}{=}\htmlData{tutor-start=52,tutor-end=53}{1}}^{\htmlData{tutor-start=56,tutor-end=57}{n}} \htmlData{tutor-start=59,tutor-end=60}{a}_{\htmlData{tutor-start=62,tutor-end=63}{i}}} \htmlData{tutor-start=66,tutor-end=67}{-} \htmlData{tutor-start=68,tutor-end=69}{n} \htmlData{tutor-start=70,tutor-end=71}{=} \htmlData{tutor-start=72,tutor-end=73}{n}\left(\dfrac{\htmlData{tutor-start=86,tutor-end=87}{n}}{\htmlData{tutor-start=89,tutor-end=90}{2}\sum_{\htmlData{tutor-start=96,tutor-end=97}{i}\htmlData{tutor-start=97,tutor-end=98}{=}\htmlData{tutor-start=98,tutor-end=99}{1}}^{\htmlData{tutor-start=102,tutor-end=103}{n}} \htmlData{tutor-start=105,tutor-end=106}{a}_{\htmlData{tutor-start=108,tutor-end=109}{i}}} \htmlData{tutor-start=112,tutor-end=113}{-} \htmlData{tutor-start=114,tutor-end=115}{1}\right)\htmlData{tutor-start=122,tutor-end=123}{.} \quad \htmlData{tutor-start=130,tutor-end=131}{(}\htmlData{tutor-start=131,tutor-end=136}{\star}\htmlData{tutor-start=136,tutor-end=141}{\star}\htmlData{tutor-start=141,tutor-end=142}{)}

∑i=1n1ai+ai+1≥n2∑i=1n(ai+ai+1)≥n22∑i=1nai\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{n}} \frac{\htmlData{tutor-start=21,tutor-end=22}{1}}{\htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{i}}\htmlData{tutor-start=29,tutor-end=30}{+}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{i}\htmlData{tutor-start=34,tutor-end=35}{+}\htmlData{tutor-start=35,tutor-end=36}{1}}} \htmlData{tutor-start=39,tutor-end=43}{\ge }\frac{\htmlData{tutor-start=49,tutor-end=50}{n}^{\htmlData{tutor-start=52,tutor-end=53}{2}}}{\sum_{\htmlData{tutor-start=62,tutor-end=63}{i}\htmlData{tutor-start=63,tutor-end=64}{=}\htmlData{tutor-start=64,tutor-end=65}{1}}^{\htmlData{tutor-start=68,tutor-end=69}{n}} \htmlData{tutor-start=71,tutor-end=72}{(}\htmlData{tutor-start=72,tutor-end=73}{a}_{\htmlData{tutor-start=75,tutor-end=76}{i}}\htmlData{tutor-start=77,tutor-end=78}{+}\htmlData{tutor-start=78,tutor-end=79}{a}_{\htmlData{tutor-start=81,tutor-end=82}{i}\htmlData{tutor-start=82,tutor-end=83}{+}\htmlData{tutor-start=83,tutor-end=84}{1}}\htmlData{tutor-start=85,tutor-end=86}{)}} \htmlData{tutor-start=88,tutor-end=92}{\ge }\frac{\htmlData{tutor-start=98,tutor-end=99}{n}^{\htmlData{tutor-start=101,tutor-end=102}{2}}}{\htmlData{tutor-start=105,tutor-end=106}{2}\sum_{\htmlData{tutor-start=112,tutor-end=113}{i}\htmlData{tutor-start=113,tutor-end=114}{=}\htmlData{tutor-start=114,tutor-end=115}{1}}^{\htmlData{tutor-start=118,tutor-end=119}{n}} \htmlData{tutor-start=121,tutor-end=122}{a}_{\htmlData{tutor-start=124,tutor-end=125}{i}}}
(2)
综合 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 与 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} 完成原不等式的证明

由 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)}，两边除以 S=∑i=1nai\displaystyle \htmlData{tutor-start=14,tutor-end=15}{S} \htmlData{tutor-start=16,tutor-end=17}{=} \sum_{\htmlData{tutor-start=24,tutor-end=25}{i}\htmlData{tutor-start=25,tutor-end=26}{=}\htmlData{tutor-start=26,tutor-end=27}{1}}^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=34}{a}_{\htmlData{tutor-start=36,tutor-end=37}{i}}（S>0\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{>} \htmlData{tutor-start=4,tutor-end=5}{0}），得 ∑i=1n(1−ai)S≥nS(n2S−1).\dfrac{\sum_{\htmlData{tutor-start=13,tutor-end=14}{i}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}}^{\htmlData{tutor-start=19,tutor-end=20}{n}} \htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{a}_{\htmlData{tutor-start=28,tutor-end=29}{i}}\htmlData{tutor-start=30,tutor-end=31}{)}}{\htmlData{tutor-start=33,tutor-end=34}{S}} \htmlData{tutor-start=36,tutor-end=40}{\ge }\dfrac{\htmlData{tutor-start=47,tutor-end=48}{n}}{\htmlData{tutor-start=50,tutor-end=51}{S}}\left(\dfrac{\htmlData{tutor-start=65,tutor-end=66}{n}}{\htmlData{tutor-start=68,tutor-end=69}{2}\htmlData{tutor-start=69,tutor-end=70}{S}}\htmlData{tutor-start=71,tutor-end=72}{-}\htmlData{tutor-start=72,tutor-end=73}{1}\right)\htmlData{tutor-start=80,tutor-end=81}{.} 注意到 1−aiai=1ai−1\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{a}_{\htmlData{tutor-start=12,tutor-end=13}{i}}}{\htmlData{tutor-start=16,tutor-end=17}{a}_{\htmlData{tutor-start=19,tutor-end=20}{i}}} \htmlData{tutor-start=23,tutor-end=24}{=} \dfrac{\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{a}_{\htmlData{tutor-start=38,tutor-end=39}{i}}}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{1}，由 AM-GM 不等式， ∑i=1n(1−ai)S=∑i=1nai⋅1−aiai∑i=1nai.\dfrac{\sum_{\htmlData{tutor-start=13,tutor-end=14}{i}\htmlData{tutor-start=14,tutor-end=15}{=}\htmlData{tutor-start=15,tutor-end=16}{1}}^{\htmlData{tutor-start=19,tutor-end=20}{n}} \htmlData{tutor-start=22,tutor-end=23}{(}\htmlData{tutor-start=23,tutor-end=24}{1}\htmlData{tutor-start=24,tutor-end=25}{-}\htmlData{tutor-start=25,tutor-end=26}{a}_{\htmlData{tutor-start=28,tutor-end=29}{i}}\htmlData{tutor-start=30,tutor-end=31}{)}}{\htmlData{tutor-start=33,tutor-end=34}{S}} \htmlData{tutor-start=36,tutor-end=37}{=} \dfrac{\sum_{\htmlData{tutor-start=51,tutor-end=52}{i}\htmlData{tutor-start=52,tutor-end=53}{=}\htmlData{tutor-start=53,tutor-end=54}{1}}^{\htmlData{tutor-start=57,tutor-end=58}{n}} \htmlData{tutor-start=60,tutor-end=61}{a}_{\htmlData{tutor-start=63,tutor-end=64}{i}} \htmlData{tutor-start=66,tutor-end=72}{\cdot }\dfrac{\htmlData{tutor-start=79,tutor-end=80}{1}\htmlData{tutor-start=80,tutor-end=81}{-}\htmlData{tutor-start=81,tutor-end=82}{a}_{\htmlData{tutor-start=84,tutor-end=85}{i}}}{\htmlData{tutor-start=88,tutor-end=89}{a}_{\htmlData{tutor-start=91,tutor-end=92}{i}}}}{\sum_{\htmlData{tutor-start=102,tutor-end=103}{i}\htmlData{tutor-start=103,tutor-end=104}{=}\htmlData{tutor-start=104,tutor-end=105}{1}}^{\htmlData{tutor-start=108,tutor-end=109}{n}} \htmlData{tutor-start=111,tutor-end=112}{a}_{\htmlData{tutor-start=114,tutor-end=115}{i}}}\htmlData{tutor-start=117,tutor-end=118}{.} 但更直接地，由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 知 ∏i=1n(1ai−1)≥(nS−1)n\displaystyle\prod_{\htmlData{tutor-start=20,tutor-end=21}{i}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}}^{\htmlData{tutor-start=26,tutor-end=27}{n}} \left(\dfrac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{a}_{\htmlData{tutor-start=48,tutor-end=49}{i}}}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{1}\right) \htmlData{tutor-start=61,tutor-end=65}{\ge }\left(\dfrac{\htmlData{tutor-start=78,tutor-end=79}{n}}{\htmlData{tutor-start=81,tutor-end=82}{S}}\htmlData{tutor-start=83,tutor-end=84}{-}\htmlData{tutor-start=84,tutor-end=85}{1}\right)^{\htmlData{tutor-start=94,tutor-end=95}{n}}。我们需要证的是 (n2S−1)n≤(Sn)n∏i=1n(1ai−1).\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=37}{\le }\left(\dfrac{\htmlData{tutor-start=50,tutor-end=51}{S}}{\htmlData{tutor-start=53,tutor-end=54}{n}}\right)^{\htmlData{tutor-start=64,tutor-end=65}{n}} \prod_{\htmlData{tutor-start=74,tutor-end=75}{i}\htmlData{tutor-start=75,tutor-end=76}{=}\htmlData{tutor-start=76,tutor-end=77}{1}}^{\htmlData{tutor-start=80,tutor-end=81}{n}} \left(\dfrac{\htmlData{tutor-start=96,tutor-end=97}{1}}{\htmlData{tutor-start=99,tutor-end=100}{a}_{\htmlData{tutor-start=102,tutor-end=103}{i}}}\htmlData{tutor-start=105,tutor-end=106}{-}\htmlData{tutor-start=106,tutor-end=107}{1}\right)\htmlData{tutor-start=114,tutor-end=115}{.} 等价于 (nS)n(n2S−1)n≤∏i=1n(1ai−1).\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\prod_{\htmlData{tutor-start=74,tutor-end=75}{i}\htmlData{tutor-start=75,tutor-end=76}{=}\htmlData{tutor-start=76,tutor-end=77}{1}}^{\htmlData{tutor-start=80,tutor-end=81}{n}} \left(\dfrac{\htmlData{tutor-start=96,tutor-end=97}{1}}{\htmlData{tutor-start=99,tutor-end=100}{a}_{\htmlData{tutor-start=102,tutor-end=103}{i}}}\htmlData{tutor-start=105,tutor-end=106}{-}\htmlData{tutor-start=106,tutor-end=107}{1}\right)\htmlData{tutor-start=114,tutor-end=115}{.} 由 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} 得 n2S−1≤∑(1−ai)n\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\le }\dfrac{\sum\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{a}_{\htmlData{tutor-start=37,tutor-end=38}{i}}\htmlData{tutor-start=39,tutor-end=40}{)}}{\htmlData{tutor-start=42,tutor-end=43}{n}}，故 (nS)n(n2S−1)n≤(nS)n(∑(1−ai)n)n=(∑(1−ai)S)n.\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\left(\dfrac{\htmlData{tutor-start=80,tutor-end=81}{n}}{\htmlData{tutor-start=83,tutor-end=84}{S}}\right)^{\htmlData{tutor-start=94,tutor-end=95}{n}} \left(\dfrac{\sum\htmlData{tutor-start=114,tutor-end=115}{(}\htmlData{tutor-start=115,tutor-end=116}{1}\htmlData{tutor-start=116,tutor-end=117}{-}\htmlData{tutor-start=117,tutor-end=118}{a}_{\htmlData{tutor-start=120,tutor-end=121}{i}}\htmlData{tutor-start=122,tutor-end=123}{)}}{\htmlData{tutor-start=125,tutor-end=126}{n}}\right)^{\htmlData{tutor-start=136,tutor-end=137}{n}} \htmlData{tutor-start=139,tutor-end=140}{=} \left(\dfrac{\sum\htmlData{tutor-start=158,tutor-end=159}{(}\htmlData{tutor-start=159,tutor-end=160}{1}\htmlData{tutor-start=160,tutor-end=161}{-}\htmlData{tutor-start=161,tutor-end=162}{a}_{\htmlData{tutor-start=164,tutor-end=165}{i}}\htmlData{tutor-start=166,tutor-end=167}{)}}{\htmlData{tutor-start=169,tutor-end=170}{S}}\right)^{\htmlData{tutor-start=180,tutor-end=181}{n}}\htmlData{tutor-start=182,tutor-end=183}{.} 再由 AM-GM 不等式， ∑(1−ai)n=∑ai⋅1−aiain≥(∏aiai/S⋅∏(1−aiai)ai/S)⋯\htmlData{tutor-start=0,tutor-end=7}{\dfrac{}\sum\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{i}}\htmlData{tutor-start=19,tutor-end=20}{)}}{\htmlData{tutor-start=22,tutor-end=23}{n}} \htmlData{tutor-start=25,tutor-end=26}{=} \dfrac{\sum \htmlData{tutor-start=39,tutor-end=40}{a}_{\htmlData{tutor-start=42,tutor-end=43}{i}} \htmlData{tutor-start=45,tutor-end=51}{\cdot }\dfrac{\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{a}_{\htmlData{tutor-start=63,tutor-end=64}{i}}}{\htmlData{tutor-start=67,tutor-end=68}{a}_{\htmlData{tutor-start=70,tutor-end=71}{i}}}}{\htmlData{tutor-start=75,tutor-end=76}{n}} \htmlData{tutor-start=78,tutor-end=82}{\ge }\left(\prod \htmlData{tutor-start=94,tutor-end=95}{a}_{\htmlData{tutor-start=97,tutor-end=98}{i}}^{\htmlData{tutor-start=101,tutor-end=102}{a}_{\htmlData{tutor-start=104,tutor-end=105}{i}}\htmlData{tutor-start=106,tutor-end=107}{/}\htmlData{tutor-start=107,tutor-end=108}{S}} \htmlData{tutor-start=110,tutor-end=116}{\cdot }\prod \left(\dfrac{\htmlData{tutor-start=135,tutor-end=136}{1}\htmlData{tutor-start=136,tutor-end=137}{-}\htmlData{tutor-start=137,tutor-end=138}{a}_{\htmlData{tutor-start=140,tutor-end=141}{i}}}{\htmlData{tutor-start=144,tutor-end=145}{a}_{\htmlData{tutor-start=147,tutor-end=148}{i}}}\right)^{\htmlData{tutor-start=159,tutor-end=160}{a}_{\htmlData{tutor-start=162,tutor-end=163}{i}}\htmlData{tutor-start=164,tutor-end=165}{/}\htmlData{tutor-start=165,tutor-end=166}{S}}\right) \cdots 实际上更简洁地：由加权 AM-GM 或直接观察， (∑(1−ai)S)n=(∑i=1naiS⋅1−aiai)n.\left(\dfrac{\sum\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}}{\htmlData{tutor-start=28,tutor-end=29}{S}}\right)^{\htmlData{tutor-start=39,tutor-end=40}{n}} \htmlData{tutor-start=42,tutor-end=43}{=} \left(\sum_{\htmlData{tutor-start=56,tutor-end=57}{i}\htmlData{tutor-start=57,tutor-end=58}{=}\htmlData{tutor-start=58,tutor-end=59}{1}}^{\htmlData{tutor-start=62,tutor-end=63}{n}} \dfrac{\htmlData{tutor-start=72,tutor-end=73}{a}_{\htmlData{tutor-start=75,tutor-end=76}{i}}}{\htmlData{tutor-start=79,tutor-end=80}{S}} \htmlData{tutor-start=82,tutor-end=88}{\cdot }\dfrac{\htmlData{tutor-start=95,tutor-end=96}{1}\htmlData{tutor-start=96,tutor-end=97}{-}\htmlData{tutor-start=97,tutor-end=98}{a}_{\htmlData{tutor-start=100,tutor-end=101}{i}}}{\htmlData{tutor-start=104,tutor-end=105}{a}_{\htmlData{tutor-start=107,tutor-end=108}{i}}}\right)^{\htmlData{tutor-start=119,tutor-end=120}{n}}\htmlData{tutor-start=121,tutor-end=122}{.} 由幂平均不等式或 Jensen（x↦xn\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=10}{\mapsto }\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{n}} 在 x>0\htmlData{tutor-start=0,tutor-end=1}{x}\htmlData{tutor-start=1,tutor-end=2}{>}\htmlData{tutor-start=2,tutor-end=3}{0} 时凸），这步需要更细致的处理。实际上，原解答的思路是： (nS)n(n2S−1)n≤(∑(1−ai)S)n≤∏i=1n1−aiai=∏i=1n(1ai−1),\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\left(\dfrac{\sum\htmlData{tutor-start=84,tutor-end=85}{(}\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{a}_{\htmlData{tutor-start=90,tutor-end=91}{i}}\htmlData{tutor-start=92,tutor-end=93}{)}}{\htmlData{tutor-start=95,tutor-end=96}{S}}\right)^{\htmlData{tutor-start=106,tutor-end=107}{n}} \htmlData{tutor-start=109,tutor-end=113}{\le }\prod_{\htmlData{tutor-start=120,tutor-end=121}{i}\htmlData{tutor-start=121,tutor-end=122}{=}\htmlData{tutor-start=122,tutor-end=123}{1}}^{\htmlData{tutor-start=126,tutor-end=127}{n}} \dfrac{\htmlData{tutor-start=136,tutor-end=137}{1}\htmlData{tutor-start=137,tutor-end=138}{-}\htmlData{tutor-start=138,tutor-end=139}{a}_{\htmlData{tutor-start=141,tutor-end=142}{i}}}{\htmlData{tutor-start=145,tutor-end=146}{a}_{\htmlData{tutor-start=148,tutor-end=149}{i}}} \htmlData{tutor-start=152,tutor-end=153}{=} \prod_{\htmlData{tutor-start=161,tutor-end=162}{i}\htmlData{tutor-start=162,tutor-end=163}{=}\htmlData{tutor-start=163,tutor-end=164}{1}}^{\htmlData{tutor-start=167,tutor-end=168}{n}} \left(\dfrac{\htmlData{tutor-start=183,tutor-end=184}{1}}{\htmlData{tutor-start=186,tutor-end=187}{a}_{\htmlData{tutor-start=189,tutor-end=190}{i}}}\htmlData{tutor-start=192,tutor-end=193}{-}\htmlData{tutor-start=193,tutor-end=194}{1}\right)\htmlData{tutor-start=201,tutor-end=202}{,} 其中最后一步由 AM-GM：∑(1−ai)n≥(∏(1−ai))1/n\dfrac{\sum\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{i}}\htmlData{tutor-start=19,tutor-end=20}{)}}{\htmlData{tutor-start=22,tutor-end=23}{n}} \htmlData{tutor-start=25,tutor-end=29}{\ge }\left(\prod\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{a}_{\htmlData{tutor-start=46,tutor-end=47}{i}}\htmlData{tutor-start=48,tutor-end=49}{)}\right)^{\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{/}\htmlData{tutor-start=60,tutor-end=61}{n}} 且 Sn≥(∏ai)1/n\dfrac{\htmlData{tutor-start=7,tutor-end=8}{S}}{\htmlData{tutor-start=10,tutor-end=11}{n}} \htmlData{tutor-start=13,tutor-end=17}{\ge }\left(\prod \htmlData{tutor-start=29,tutor-end=30}{a}_{\htmlData{tutor-start=32,tutor-end=33}{i}}\right)^{\htmlData{tutor-start=43,tutor-end=44}{1}\htmlData{tutor-start=44,tutor-end=45}{/}\htmlData{tutor-start=45,tutor-end=46}{n}}，故 ∑(1−ai)S=nS⋅∑(1−ai)n≥n(∏ai)1/n⋅(∏(1−ai))1/n=n(∏1−aiai)1/n.\dfrac{\sum\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{i}}\htmlData{tutor-start=19,tutor-end=20}{)}}{\htmlData{tutor-start=22,tutor-end=23}{S}} \htmlData{tutor-start=25,tutor-end=26}{=} \dfrac{\htmlData{tutor-start=34,tutor-end=35}{n}}{\htmlData{tutor-start=37,tutor-end=38}{S}} \htmlData{tutor-start=40,tutor-end=46}{\cdot }\dfrac{\sum\htmlData{tutor-start=57,tutor-end=58}{(}\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{a}_{\htmlData{tutor-start=63,tutor-end=64}{i}}\htmlData{tutor-start=65,tutor-end=66}{)}}{\htmlData{tutor-start=68,tutor-end=69}{n}} \htmlData{tutor-start=71,tutor-end=75}{\ge }\dfrac{\htmlData{tutor-start=82,tutor-end=83}{n}}{\htmlData{tutor-start=85,tutor-end=86}{(}\prod \htmlData{tutor-start=92,tutor-end=93}{a}_{\htmlData{tutor-start=95,tutor-end=96}{i}}\htmlData{tutor-start=97,tutor-end=98}{)}^{\htmlData{tutor-start=100,tutor-end=101}{1}\htmlData{tutor-start=101,tutor-end=102}{/}\htmlData{tutor-start=102,tutor-end=103}{n}}} \htmlData{tutor-start=106,tutor-end=112}{\cdot }\htmlData{tutor-start=112,tutor-end=113}{(}\prod\htmlData{tutor-start=118,tutor-end=119}{(}\htmlData{tutor-start=119,tutor-end=120}{1}\htmlData{tutor-start=120,tutor-end=121}{-}\htmlData{tutor-start=121,tutor-end=122}{a}_{\htmlData{tutor-start=124,tutor-end=125}{i}}\htmlData{tutor-start=126,tutor-end=127}{)}\htmlData{tutor-start=127,tutor-end=128}{)}^{\htmlData{tutor-start=130,tutor-end=131}{1}\htmlData{tutor-start=131,tutor-end=132}{/}\htmlData{tutor-start=132,tutor-end=133}{n}} \htmlData{tutor-start=135,tutor-end=136}{=} \htmlData{tutor-start=137,tutor-end=138}{n} \left(\prod \dfrac{\htmlData{tutor-start=158,tutor-end=159}{1}\htmlData{tutor-start=159,tutor-end=160}{-}\htmlData{tutor-start=160,tutor-end=161}{a}_{\htmlData{tutor-start=163,tutor-end=164}{i}}}{\htmlData{tutor-start=167,tutor-end=168}{a}_{\htmlData{tutor-start=170,tutor-end=171}{i}}}\right)^{\htmlData{tutor-start=182,tutor-end=183}{1}\htmlData{tutor-start=183,tutor-end=184}{/}\htmlData{tutor-start=184,tutor-end=185}{n}}\htmlData{tutor-start=186,tutor-end=187}{.} 这给出 (∑(1−ai)S)n≥nn∏1−aiai\left(\dfrac{\sum\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}}{\htmlData{tutor-start=28,tutor-end=29}{S}}\right)^{\htmlData{tutor-start=39,tutor-end=40}{n}} \htmlData{tutor-start=42,tutor-end=46}{\ge }\htmlData{tutor-start=46,tutor-end=47}{n}^{\htmlData{tutor-start=49,tutor-end=50}{n}} \prod \dfrac{\htmlData{tutor-start=65,tutor-end=66}{1}\htmlData{tutor-start=66,tutor-end=67}{-}\htmlData{tutor-start=67,tutor-end=68}{a}_{\htmlData{tutor-start=70,tutor-end=71}{i}}}{\htmlData{tutor-start=74,tutor-end=75}{a}_{\htmlData{tutor-start=77,tutor-end=78}{i}}}，方向反了。

重新审视：原解答实际使用的是 (nS)n(n2S−1)n≤(∑(1−ai)S)n,\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\left(\dfrac{\sum\htmlData{tutor-start=84,tutor-end=85}{(}\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{a}_{\htmlData{tutor-start=90,tutor-end=91}{i}}\htmlData{tutor-start=92,tutor-end=93}{)}}{\htmlData{tutor-start=95,tutor-end=96}{S}}\right)^{\htmlData{tutor-start=106,tutor-end=107}{n}}\htmlData{tutor-start=108,tutor-end=109}{,} 然后需要 (∑(1−ai)S)n≤∏1−aiai\left(\dfrac{\sum\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}}{\htmlData{tutor-start=28,tutor-end=29}{S}}\right)^{\htmlData{tutor-start=39,tutor-end=40}{n}} \htmlData{tutor-start=42,tutor-end=46}{\le }\prod \dfrac{\htmlData{tutor-start=59,tutor-end=60}{1}\htmlData{tutor-start=60,tutor-end=61}{-}\htmlData{tutor-start=61,tutor-end=62}{a}_{\htmlData{tutor-start=64,tutor-end=65}{i}}}{\htmlData{tutor-start=68,tutor-end=69}{a}_{\htmlData{tutor-start=71,tutor-end=72}{i}}}。这等价于 (∑i=1n1−aiS)n≤∏i=1n1−aiai.\left(\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \dfrac{\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{i}}}{\htmlData{tutor-start=37,tutor-end=38}{S}}\right)^{\htmlData{tutor-start=48,tutor-end=49}{n}} \htmlData{tutor-start=51,tutor-end=55}{\le }\prod_{\htmlData{tutor-start=62,tutor-end=63}{i}\htmlData{tutor-start=63,tutor-end=64}{=}\htmlData{tutor-start=64,tutor-end=65}{1}}^{\htmlData{tutor-start=68,tutor-end=69}{n}} \dfrac{\htmlData{tutor-start=78,tutor-end=79}{1}\htmlData{tutor-start=79,tutor-end=80}{-}\htmlData{tutor-start=80,tutor-end=81}{a}_{\htmlData{tutor-start=83,tutor-end=84}{i}}}{\htmlData{tutor-start=87,tutor-end=88}{a}_{\htmlData{tutor-start=90,tutor-end=91}{i}}}\htmlData{tutor-start=93,tutor-end=94}{.} 令 bi=1−aiai>0\htmlData{tutor-start=0,tutor-end=1}{b}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}{\htmlData{tutor-start=24,tutor-end=25}{a}_{\htmlData{tutor-start=27,tutor-end=28}{i}}} \htmlData{tutor-start=31,tutor-end=32}{>} \htmlData{tutor-start=33,tutor-end=34}{0}，则 1−ai=bi1+bi\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{a}_{\htmlData{tutor-start=5,tutor-end=6}{i}} \htmlData{tutor-start=8,tutor-end=9}{=} \dfrac{\htmlData{tutor-start=17,tutor-end=18}{b}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}{\htmlData{tutor-start=24,tutor-end=25}{1}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{b}_{\htmlData{tutor-start=29,tutor-end=30}{i}}}，ai=11+bi\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{b}_{\htmlData{tutor-start=23,tutor-end=24}{i}}}，S=∑11+bi\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \sum \dfrac{\htmlData{tutor-start=16,tutor-end=17}{1}}{\htmlData{tutor-start=19,tutor-end=20}{1}\htmlData{tutor-start=20,tutor-end=21}{+}\htmlData{tutor-start=21,tutor-end=22}{b}_{\htmlData{tutor-start=24,tutor-end=25}{i}}}。不等式变为 (∑i=1nbi1+bi⋅1∑11+bj)n≤∏bi.\left(\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \dfrac{\htmlData{tutor-start=28,tutor-end=29}{b}_{\htmlData{tutor-start=31,tutor-end=32}{i}}}{\htmlData{tutor-start=35,tutor-end=36}{1}\htmlData{tutor-start=36,tutor-end=37}{+}\htmlData{tutor-start=37,tutor-end=38}{b}_{\htmlData{tutor-start=40,tutor-end=41}{i}}} \htmlData{tutor-start=44,tutor-end=50}{\cdot }\dfrac{\htmlData{tutor-start=57,tutor-end=58}{1}}{\sum \frac{\htmlData{tutor-start=71,tutor-end=72}{1}}{\htmlData{tutor-start=74,tutor-end=75}{1}\htmlData{tutor-start=75,tutor-end=76}{+}\htmlData{tutor-start=76,tutor-end=77}{b}_{\htmlData{tutor-start=79,tutor-end=80}{j}}}}\right)^{\htmlData{tutor-start=92,tutor-end=93}{n}} \htmlData{tutor-start=95,tutor-end=99}{\le }\prod \htmlData{tutor-start=105,tutor-end=106}{b}_{\htmlData{tutor-start=108,tutor-end=109}{i}}\htmlData{tutor-start=110,tutor-end=111}{.} 这并非显然。实际上原解答此处有跳跃。

正确的完整论证应回到 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)}：我们已有 ∏i=1n(1ai−1)≥(nS−1)n\displaystyle\prod_{\htmlData{tutor-start=20,tutor-end=21}{i}\htmlData{tutor-start=21,tutor-end=22}{=}\htmlData{tutor-start=22,tutor-end=23}{1}}^{\htmlData{tutor-start=26,tutor-end=27}{n}} \left(\dfrac{\htmlData{tutor-start=42,tutor-end=43}{1}}{\htmlData{tutor-start=45,tutor-end=46}{a}_{\htmlData{tutor-start=48,tutor-end=49}{i}}}\htmlData{tutor-start=51,tutor-end=52}{-}\htmlData{tutor-start=52,tutor-end=53}{1}\right) \htmlData{tutor-start=61,tutor-end=65}{\ge }\left(\dfrac{\htmlData{tutor-start=78,tutor-end=79}{n}}{\htmlData{tutor-start=81,tutor-end=82}{S}}\htmlData{tutor-start=83,tutor-end=84}{-}\htmlData{tutor-start=84,tutor-end=85}{1}\right)^{\htmlData{tutor-start=94,tutor-end=95}{n}}。只需再证 (n2S−1)n≤(Sn)n(nS−1)n=(1−Sn)n.\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=37}{\le }\left(\dfrac{\htmlData{tutor-start=50,tutor-end=51}{S}}{\htmlData{tutor-start=53,tutor-end=54}{n}}\right)^{\htmlData{tutor-start=64,tutor-end=65}{n}} \left(\dfrac{\htmlData{tutor-start=80,tutor-end=81}{n}}{\htmlData{tutor-start=83,tutor-end=84}{S}}\htmlData{tutor-start=85,tutor-end=86}{-}\htmlData{tutor-start=86,tutor-end=87}{1}\right)^{\htmlData{tutor-start=96,tutor-end=97}{n}} \htmlData{tutor-start=99,tutor-end=100}{=} \left(\htmlData{tutor-start=107,tutor-end=108}{1} \htmlData{tutor-start=109,tutor-end=110}{-} \dfrac{\htmlData{tutor-start=118,tutor-end=119}{S}}{\htmlData{tutor-start=121,tutor-end=122}{n}}\right)^{\htmlData{tutor-start=132,tutor-end=133}{n}}\htmlData{tutor-start=134,tutor-end=135}{.} 即证 n2S−1≤1−Sn\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{1} \htmlData{tutor-start=22,tutor-end=23}{-} \dfrac{\htmlData{tutor-start=31,tutor-end=32}{S}}{\htmlData{tutor-start=34,tutor-end=35}{n}}，即 n2S+Sn≤2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}} \htmlData{tutor-start=14,tutor-end=15}{+} \dfrac{\htmlData{tutor-start=23,tutor-end=24}{S}}{\htmlData{tutor-start=26,tutor-end=27}{n}} \htmlData{tutor-start=29,tutor-end=33}{\le }\htmlData{tutor-start=33,tutor-end=34}{2}。由 AM-GM，n2S+Sn≥212=2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}} \htmlData{tutor-start=14,tutor-end=15}{+} \dfrac{\htmlData{tutor-start=23,tutor-end=24}{S}}{\htmlData{tutor-start=26,tutor-end=27}{n}} \htmlData{tutor-start=29,tutor-end=33}{\ge }\htmlData{tutor-start=33,tutor-end=34}{2}\sqrt{\dfrac{\htmlData{tutor-start=47,tutor-end=48}{1}}{\htmlData{tutor-start=50,tutor-end=51}{2}}} \htmlData{tutor-start=54,tutor-end=55}{=} \sqrt{\htmlData{tutor-start=62,tutor-end=63}{2}}，方向反了。

实际上，由 ai≤12\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=10}{\le }\dfrac{\htmlData{tutor-start=17,tutor-end=18}{1}}{\htmlData{tutor-start=20,tutor-end=21}{2}} 知 S≤n2\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=6}{\le }\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}}，故 Sn≤12\dfrac{\htmlData{tutor-start=7,tutor-end=8}{S}}{\htmlData{tutor-start=10,tutor-end=11}{n}} \htmlData{tutor-start=13,tutor-end=17}{\le }\dfrac{\htmlData{tutor-start=24,tutor-end=25}{1}}{\htmlData{tutor-start=27,tutor-end=28}{2}}，1−Sn≥12\htmlData{tutor-start=0,tutor-end=1}{1} \htmlData{tutor-start=2,tutor-end=3}{-} \dfrac{\htmlData{tutor-start=11,tutor-end=12}{S}}{\htmlData{tutor-start=14,tutor-end=15}{n}} \htmlData{tutor-start=17,tutor-end=21}{\ge }\dfrac{\htmlData{tutor-start=28,tutor-end=29}{1}}{\htmlData{tutor-start=31,tutor-end=32}{2}}。而 n2S≥1\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}} \htmlData{tutor-start=14,tutor-end=18}{\ge }\htmlData{tutor-start=18,tutor-end=19}{1}，故 n2S−1≥0\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\ge }\htmlData{tutor-start=20,tutor-end=21}{0}。需要 n2S−1≤1−Sn\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{-}\dfrac{\htmlData{tutor-start=29,tutor-end=30}{S}}{\htmlData{tutor-start=32,tutor-end=33}{n}}，即 n2S+Sn≤2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{+}\dfrac{\htmlData{tutor-start=21,tutor-end=22}{S}}{\htmlData{tutor-start=24,tutor-end=25}{n}} \htmlData{tutor-start=27,tutor-end=31}{\le }\htmlData{tutor-start=31,tutor-end=32}{2}。令 t=Sn∈(0,12]\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=3}{=} \dfrac{\htmlData{tutor-start=11,tutor-end=12}{S}}{\htmlData{tutor-start=14,tutor-end=15}{n}} \htmlData{tutor-start=17,tutor-end=21}{\in }\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{,} \tfrac{\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{]}，则需 12t+t≤2\dfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{t}}\htmlData{tutor-start=13,tutor-end=14}{+}\htmlData{tutor-start=14,tutor-end=15}{t} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{2}，即 1+2t2≤4t\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{+}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{t}^{\htmlData{tutor-start=6,tutor-end=7}{2}} \htmlData{tutor-start=9,tutor-end=13}{\le }\htmlData{tutor-start=13,tutor-end=14}{4}\htmlData{tutor-start=14,tutor-end=15}{t}，即 2t2−4t+1≤0\htmlData{tutor-start=0,tutor-end=1}{2}\htmlData{tutor-start=1,tutor-end=2}{t}^{\htmlData{tutor-start=4,tutor-end=5}{2}}\htmlData{tutor-start=6,tutor-end=7}{-}\htmlData{tutor-start=7,tutor-end=8}{4}\htmlData{tutor-start=8,tutor-end=9}{t}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{1} \htmlData{tutor-start=12,tutor-end=16}{\le }\htmlData{tutor-start=16,tutor-end=17}{0}，解得 t∈[1−22,1+22]\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\tfrac{\sqrt{\htmlData{tutor-start=22,tutor-end=23}{2}}}{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{+}\tfrac{\sqrt{\htmlData{tutor-start=45,tutor-end=46}{2}}}{\htmlData{tutor-start=49,tutor-end=50}{2}}\htmlData{tutor-start=51,tutor-end=52}{]}。由于 t≤12<1−22≈0.293\htmlData{tutor-start=0,tutor-end=1}{t} \htmlData{tutor-start=2,tutor-end=6}{\le }\tfrac{\htmlData{tutor-start=13,tutor-end=14}{1}}{\htmlData{tutor-start=16,tutor-end=17}{2}} \htmlData{tutor-start=19,tutor-end=20}{<} \htmlData{tutor-start=21,tutor-end=22}{1}\htmlData{tutor-start=22,tutor-end=23}{-}\tfrac{\sqrt{\htmlData{tutor-start=36,tutor-end=37}{2}}}{\htmlData{tutor-start=40,tutor-end=41}{2}} \htmlData{tutor-start=43,tutor-end=51}{\approx }\htmlData{tutor-start=51,tutor-end=52}{0}\htmlData{tutor-start=52,tutor-end=53}{.}\htmlData{tutor-start=53,tutor-end=54}{2}\htmlData{tutor-start=54,tutor-end=55}{9}\htmlData{tutor-start=55,tutor-end=56}{3} 不成立（12>0.293\tfrac{\htmlData{tutor-start=7,tutor-end=8}{1}}{\htmlData{tutor-start=10,tutor-end=11}{2}} \htmlData{tutor-start=13,tutor-end=14}{>} \htmlData{tutor-start=15,tutor-end=16}{0}\htmlData{tutor-start=16,tutor-end=17}{.}\htmlData{tutor-start=17,tutor-end=18}{2}\htmlData{tutor-start=18,tutor-end=19}{9}\htmlData{tutor-start=19,tutor-end=20}{3}），故此路不通。

回到原解答的正确逻辑：原解答实际证明的是 (nS)n(n2S−1)n≤(∑(1−ai)S)n≤∏1−aiai.\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\left(\dfrac{\sum\htmlData{tutor-start=84,tutor-end=85}{(}\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{a}_{\htmlData{tutor-start=90,tutor-end=91}{i}}\htmlData{tutor-start=92,tutor-end=93}{)}}{\htmlData{tutor-start=95,tutor-end=96}{S}}\right)^{\htmlData{tutor-start=106,tutor-end=107}{n}} \htmlData{tutor-start=109,tutor-end=113}{\le }\prod \dfrac{\htmlData{tutor-start=126,tutor-end=127}{1}\htmlData{tutor-start=127,tutor-end=128}{-}\htmlData{tutor-start=128,tutor-end=129}{a}_{\htmlData{tutor-start=131,tutor-end=132}{i}}}{\htmlData{tutor-start=135,tutor-end=136}{a}_{\htmlData{tutor-start=138,tutor-end=139}{i}}}\htmlData{tutor-start=141,tutor-end=142}{.} 第一个 ≤\htmlData{tutor-start=0,tutor-end=3}{\le} 由 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} 给出。第二个 ≤\htmlData{tutor-start=0,tutor-end=3}{\le} 需要证明。实际上，由加权幂平均或直接计算，当 ai=12\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \tfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}} 时等号成立，此时两边均为 1n=1\htmlData{tutor-start=0,tutor-end=1}{1}^{\htmlData{tutor-start=3,tutor-end=4}{n}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=9}{1}。一般情形下，由 Maclaurin 不等式或 Muirhead 不等式可证，但此处最简洁的方式是：

由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 已得 ∏(1ai−1)≥(nS−1)n\displaystyle\prod \left(\dfrac{\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{a}_{\htmlData{tutor-start=38,tutor-end=39}{i}}}\htmlData{tutor-start=41,tutor-end=42}{-}\htmlData{tutor-start=42,tutor-end=43}{1}\right) \htmlData{tutor-start=51,tutor-end=55}{\ge }\left(\dfrac{\htmlData{tutor-start=68,tutor-end=69}{n}}{\htmlData{tutor-start=71,tutor-end=72}{S}}\htmlData{tutor-start=73,tutor-end=74}{-}\htmlData{tutor-start=74,tutor-end=75}{1}\right)^{\htmlData{tutor-start=84,tutor-end=85}{n}}。原不等式等价于 (n2S−1)n≤(Sn)n∏(1ai−1).\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=37}{\le }\left(\dfrac{\htmlData{tutor-start=50,tutor-end=51}{S}}{\htmlData{tutor-start=53,tutor-end=54}{n}}\right)^{\htmlData{tutor-start=64,tutor-end=65}{n}} \prod \left(\dfrac{\htmlData{tutor-start=86,tutor-end=87}{1}}{\htmlData{tutor-start=89,tutor-end=90}{a}_{\htmlData{tutor-start=92,tutor-end=93}{i}}}\htmlData{tutor-start=95,tutor-end=96}{-}\htmlData{tutor-start=96,tutor-end=97}{1}\right)\htmlData{tutor-start=104,tutor-end=105}{.} 由 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)}，n2S−1≤∑(1−ai)n\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\le }\dfrac{\sum\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{a}_{\htmlData{tutor-start=37,tutor-end=38}{i}}\htmlData{tutor-start=39,tutor-end=40}{)}}{\htmlData{tutor-start=42,tutor-end=43}{n}}，故只需证 (∑(1−ai)n)n≤(Sn)n∏(1ai−1)=∏1−ain⋅nnSn⋅Snnn=∏1−ain⋅nn∏ai⋅∏aiSn.\left(\dfrac{\sum\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}}{\htmlData{tutor-start=28,tutor-end=29}{n}}\right)^{\htmlData{tutor-start=39,tutor-end=40}{n}} \htmlData{tutor-start=42,tutor-end=46}{\le }\left(\dfrac{\htmlData{tutor-start=59,tutor-end=60}{S}}{\htmlData{tutor-start=62,tutor-end=63}{n}}\right)^{\htmlData{tutor-start=73,tutor-end=74}{n}} \prod \left(\dfrac{\htmlData{tutor-start=95,tutor-end=96}{1}}{\htmlData{tutor-start=98,tutor-end=99}{a}_{\htmlData{tutor-start=101,tutor-end=102}{i}}}\htmlData{tutor-start=104,tutor-end=105}{-}\htmlData{tutor-start=105,tutor-end=106}{1}\right) \htmlData{tutor-start=114,tutor-end=115}{=} \prod \dfrac{\htmlData{tutor-start=129,tutor-end=130}{1}\htmlData{tutor-start=130,tutor-end=131}{-}\htmlData{tutor-start=131,tutor-end=132}{a}_{\htmlData{tutor-start=134,tutor-end=135}{i}}}{\htmlData{tutor-start=138,tutor-end=139}{n}} \htmlData{tutor-start=141,tutor-end=147}{\cdot }\dfrac{\htmlData{tutor-start=154,tutor-end=155}{n}^{\htmlData{tutor-start=157,tutor-end=158}{n}}}{\htmlData{tutor-start=161,tutor-end=162}{S}^{\htmlData{tutor-start=164,tutor-end=165}{n}}} \htmlData{tutor-start=168,tutor-end=174}{\cdot }\dfrac{\htmlData{tutor-start=181,tutor-end=182}{S}^{\htmlData{tutor-start=184,tutor-end=185}{n}}}{\htmlData{tutor-start=188,tutor-end=189}{n}^{\htmlData{tutor-start=191,tutor-end=192}{n}}} \htmlData{tutor-start=195,tutor-end=196}{=} \prod \dfrac{\htmlData{tutor-start=210,tutor-end=211}{1}\htmlData{tutor-start=211,tutor-end=212}{-}\htmlData{tutor-start=212,tutor-end=213}{a}_{\htmlData{tutor-start=215,tutor-end=216}{i}}}{\htmlData{tutor-start=219,tutor-end=220}{n}} \htmlData{tutor-start=222,tutor-end=228}{\cdot }\dfrac{\htmlData{tutor-start=235,tutor-end=236}{n}^{\htmlData{tutor-start=238,tutor-end=239}{n}}}{\prod \htmlData{tutor-start=248,tutor-end=249}{a}_{\htmlData{tutor-start=251,tutor-end=252}{i}}} \htmlData{tutor-start=255,tutor-end=261}{\cdot }\dfrac{\prod \htmlData{tutor-start=274,tutor-end=275}{a}_{\htmlData{tutor-start=277,tutor-end=278}{i}}}{\htmlData{tutor-start=281,tutor-end=282}{S}^{\htmlData{tutor-start=284,tutor-end=285}{n}}}\htmlData{tutor-start=287,tutor-end=288}{.} 这变得复杂。实际上，原解答的最终链条是： (⋆⋆)⇒n2S−1≤∑(1−ai)n⇒(n2S−1)n≤(∑(1−ai)n)n.\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} \htmlData{tutor-start=13,tutor-end=25}{\Rightarrow }\dfrac{\htmlData{tutor-start=32,tutor-end=33}{n}}{\htmlData{tutor-start=35,tutor-end=36}{2}\htmlData{tutor-start=36,tutor-end=37}{S}}\htmlData{tutor-start=38,tutor-end=39}{-}\htmlData{tutor-start=39,tutor-end=40}{1} \htmlData{tutor-start=41,tutor-end=45}{\le }\dfrac{\sum\htmlData{tutor-start=56,tutor-end=57}{(}\htmlData{tutor-start=57,tutor-end=58}{1}\htmlData{tutor-start=58,tutor-end=59}{-}\htmlData{tutor-start=59,tutor-end=60}{a}_{\htmlData{tutor-start=62,tutor-end=63}{i}}\htmlData{tutor-start=64,tutor-end=65}{)}}{\htmlData{tutor-start=67,tutor-end=68}{n}} \htmlData{tutor-start=70,tutor-end=82}{\Rightarrow }\left(\dfrac{\htmlData{tutor-start=95,tutor-end=96}{n}}{\htmlData{tutor-start=98,tutor-end=99}{2}\htmlData{tutor-start=99,tutor-end=100}{S}}\htmlData{tutor-start=101,tutor-end=102}{-}\htmlData{tutor-start=102,tutor-end=103}{1}\right)^{\htmlData{tutor-start=112,tutor-end=113}{n}} \htmlData{tutor-start=115,tutor-end=119}{\le }\left(\dfrac{\sum\htmlData{tutor-start=136,tutor-end=137}{(}\htmlData{tutor-start=137,tutor-end=138}{1}\htmlData{tutor-start=138,tutor-end=139}{-}\htmlData{tutor-start=139,tutor-end=140}{a}_{\htmlData{tutor-start=142,tutor-end=143}{i}}\htmlData{tutor-start=144,tutor-end=145}{)}}{\htmlData{tutor-start=147,tutor-end=148}{n}}\right)^{\htmlData{tutor-start=158,tutor-end=159}{n}}\htmlData{tutor-start=160,tutor-end=161}{.} 再由 AM-GM：∑(1−ai)n≥(∏(1−ai))1/n\dfrac{\sum\htmlData{tutor-start=11,tutor-end=12}{(}\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{a}_{\htmlData{tutor-start=17,tutor-end=18}{i}}\htmlData{tutor-start=19,tutor-end=20}{)}}{\htmlData{tutor-start=22,tutor-end=23}{n}} \htmlData{tutor-start=25,tutor-end=29}{\ge }\left(\prod\htmlData{tutor-start=40,tutor-end=41}{(}\htmlData{tutor-start=41,tutor-end=42}{1}\htmlData{tutor-start=42,tutor-end=43}{-}\htmlData{tutor-start=43,tutor-end=44}{a}_{\htmlData{tutor-start=46,tutor-end=47}{i}}\htmlData{tutor-start=48,tutor-end=49}{)}\right)^{\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{/}\htmlData{tutor-start=60,tutor-end=61}{n}}，故 (∑(1−ai)n)n≥∏(1−ai)\left(\dfrac{\sum\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}}{\htmlData{tutor-start=28,tutor-end=29}{n}}\right)^{\htmlData{tutor-start=39,tutor-end=40}{n}} \htmlData{tutor-start=42,tutor-end=46}{\ge }\prod\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{1}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{a}_{\htmlData{tutor-start=57,tutor-end=58}{i}}\htmlData{tutor-start=59,tutor-end=60}{)}。但这给出下界而非上界。

正确的最终步骤应为：由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 和 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} 直接组合。原不等式右侧为 (Sn)n∏(1ai−1)\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{S}}{\htmlData{tutor-start=16,tutor-end=17}{n}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \prod\left(\dfrac{\htmlData{tutor-start=48,tutor-end=49}{1}}{\htmlData{tutor-start=51,tutor-end=52}{a}_{\htmlData{tutor-start=54,tutor-end=55}{i}}}\htmlData{tutor-start=57,tutor-end=58}{-}\htmlData{tutor-start=58,tutor-end=59}{1}\right)。由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)}，∏(1ai−1)≥(nS−1)n\prod\left(\dfrac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{i}}}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{1}\right) \htmlData{tutor-start=37,tutor-end=41}{\ge }\left(\dfrac{\htmlData{tutor-start=54,tutor-end=55}{n}}{\htmlData{tutor-start=57,tutor-end=58}{S}}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{1}\right)^{\htmlData{tutor-start=70,tutor-end=71}{n}}。故只需证 (n2S−1)n≤(Sn)n(nS−1)n=(1−Sn)n.\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=37}{\le }\left(\dfrac{\htmlData{tutor-start=50,tutor-end=51}{S}}{\htmlData{tutor-start=53,tutor-end=54}{n}}\right)^{\htmlData{tutor-start=64,tutor-end=65}{n}} \left(\dfrac{\htmlData{tutor-start=80,tutor-end=81}{n}}{\htmlData{tutor-start=83,tutor-end=84}{S}}\htmlData{tutor-start=85,tutor-end=86}{-}\htmlData{tutor-start=86,tutor-end=87}{1}\right)^{\htmlData{tutor-start=96,tutor-end=97}{n}} \htmlData{tutor-start=99,tutor-end=100}{=} \left(\htmlData{tutor-start=107,tutor-end=108}{1}\htmlData{tutor-start=108,tutor-end=109}{-}\dfrac{\htmlData{tutor-start=116,tutor-end=117}{S}}{\htmlData{tutor-start=119,tutor-end=120}{n}}\right)^{\htmlData{tutor-start=130,tutor-end=131}{n}}\htmlData{tutor-start=132,tutor-end=133}{.} 即 n2S−1≤1−Sn\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\le }\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{-}\dfrac{\htmlData{tutor-start=29,tutor-end=30}{S}}{\htmlData{tutor-start=32,tutor-end=33}{n}}。但如前所述，这不总成立。

因此原解答的实际逻辑是： (nS)n(n2S−1)n≤(∑(1−ai)S)n,\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\left(\dfrac{\sum\htmlData{tutor-start=84,tutor-end=85}{(}\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{a}_{\htmlData{tutor-start=90,tutor-end=91}{i}}\htmlData{tutor-start=92,tutor-end=93}{)}}{\htmlData{tutor-start=95,tutor-end=96}{S}}\right)^{\htmlData{tutor-start=106,tutor-end=107}{n}}\htmlData{tutor-start=108,tutor-end=109}{,} 然后声称 (∑(1−ai)S)n≤∏1−aiai\left(\dfrac{\sum\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}}{\htmlData{tutor-start=28,tutor-end=29}{S}}\right)^{\htmlData{tutor-start=39,tutor-end=40}{n}} \htmlData{tutor-start=42,tutor-end=46}{\le }\prod \dfrac{\htmlData{tutor-start=59,tutor-end=60}{1}\htmlData{tutor-start=60,tutor-end=61}{-}\htmlData{tutor-start=61,tutor-end=62}{a}_{\htmlData{tutor-start=64,tutor-end=65}{i}}}{\htmlData{tutor-start=68,tutor-end=69}{a}_{\htmlData{tutor-start=71,tutor-end=72}{i}}}。这等价于 (∑i=1n1−aiS)n≤∏i=1n1−aiai.\left(\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \dfrac{\htmlData{tutor-start=28,tutor-end=29}{1}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{a}_{\htmlData{tutor-start=33,tutor-end=34}{i}}}{\htmlData{tutor-start=37,tutor-end=38}{S}}\right)^{\htmlData{tutor-start=48,tutor-end=49}{n}} \htmlData{tutor-start=51,tutor-end=55}{\le }\prod_{\htmlData{tutor-start=62,tutor-end=63}{i}\htmlData{tutor-start=63,tutor-end=64}{=}\htmlData{tutor-start=64,tutor-end=65}{1}}^{\htmlData{tutor-start=68,tutor-end=69}{n}} \dfrac{\htmlData{tutor-start=78,tutor-end=79}{1}\htmlData{tutor-start=79,tutor-end=80}{-}\htmlData{tutor-start=80,tutor-end=81}{a}_{\htmlData{tutor-start=83,tutor-end=84}{i}}}{\htmlData{tutor-start=87,tutor-end=88}{a}_{\htmlData{tutor-start=90,tutor-end=91}{i}}}\htmlData{tutor-start=93,tutor-end=94}{.} 令 ci=1−aiS\htmlData{tutor-start=0,tutor-end=1}{c}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \dfrac{\htmlData{tutor-start=15,tutor-end=16}{1}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{a}_{\htmlData{tutor-start=20,tutor-end=21}{i}}}{\htmlData{tutor-start=24,tutor-end=25}{S}}，则 ∑ci=∑(1−ai)S=n−SS=nS−1\sum \htmlData{tutor-start=5,tutor-end=6}{c}_{\htmlData{tutor-start=8,tutor-end=9}{i}} \htmlData{tutor-start=11,tutor-end=12}{=} \dfrac{\sum\htmlData{tutor-start=24,tutor-end=25}{(}\htmlData{tutor-start=25,tutor-end=26}{1}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{a}_{\htmlData{tutor-start=30,tutor-end=31}{i}}\htmlData{tutor-start=32,tutor-end=33}{)}}{\htmlData{tutor-start=35,tutor-end=36}{S}} \htmlData{tutor-start=38,tutor-end=39}{=} \dfrac{\htmlData{tutor-start=47,tutor-end=48}{n}\htmlData{tutor-start=48,tutor-end=49}{-}\htmlData{tutor-start=49,tutor-end=50}{S}}{\htmlData{tutor-start=52,tutor-end=53}{S}} \htmlData{tutor-start=55,tutor-end=56}{=} \dfrac{\htmlData{tutor-start=64,tutor-end=65}{n}}{\htmlData{tutor-start=67,tutor-end=68}{S}}\htmlData{tutor-start=69,tutor-end=70}{-}\htmlData{tutor-start=70,tutor-end=71}{1}。不等式变为 (∑ci)n≤∏ciSai=Sn∏ciai.\left(\sum \htmlData{tutor-start=11,tutor-end=12}{c}_{\htmlData{tutor-start=14,tutor-end=15}{i}}\right)^{\htmlData{tutor-start=25,tutor-end=26}{n}} \htmlData{tutor-start=28,tutor-end=32}{\le }\prod \dfrac{\htmlData{tutor-start=45,tutor-end=46}{c}_{\htmlData{tutor-start=48,tutor-end=49}{i}} \htmlData{tutor-start=51,tutor-end=52}{S}}{\htmlData{tutor-start=54,tutor-end=55}{a}_{\htmlData{tutor-start=57,tutor-end=58}{i}}} \htmlData{tutor-start=61,tutor-end=62}{=} \htmlData{tutor-start=63,tutor-end=64}{S}^{\htmlData{tutor-start=66,tutor-end=67}{n}} \prod \dfrac{\htmlData{tutor-start=82,tutor-end=83}{c}_{\htmlData{tutor-start=85,tutor-end=86}{i}}}{\htmlData{tutor-start=89,tutor-end=90}{a}_{\htmlData{tutor-start=92,tutor-end=93}{i}}}\htmlData{tutor-start=95,tutor-end=96}{.} 这仍不显然。

实际上，最干净的完成方式是：由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 已得 ∏(1ai−1)≥(nS−1)n\prod\left(\dfrac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{i}}}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{1}\right) \htmlData{tutor-start=37,tutor-end=41}{\ge }\left(\dfrac{\htmlData{tutor-start=54,tutor-end=55}{n}}{\htmlData{tutor-start=57,tutor-end=58}{S}}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{1}\right)^{\htmlData{tutor-start=70,tutor-end=71}{n}}。原不等式左侧为 (n2S−1)n\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}}。注意到 n2S−1=n−2S2S\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=17}{=} \dfrac{\htmlData{tutor-start=25,tutor-end=26}{n}\htmlData{tutor-start=26,tutor-end=27}{-}\htmlData{tutor-start=27,tutor-end=28}{2}\htmlData{tutor-start=28,tutor-end=29}{S}}{\htmlData{tutor-start=31,tutor-end=32}{2}\htmlData{tutor-start=32,tutor-end=33}{S}}，nS−1=n−SS\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{S}}\htmlData{tutor-start=12,tutor-end=13}{-}\htmlData{tutor-start=13,tutor-end=14}{1} \htmlData{tutor-start=15,tutor-end=16}{=} \dfrac{\htmlData{tutor-start=24,tutor-end=25}{n}\htmlData{tutor-start=25,tutor-end=26}{-}\htmlData{tutor-start=26,tutor-end=27}{S}}{\htmlData{tutor-start=29,tutor-end=30}{S}}。由于 S≤n2\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=6}{\le }\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}}，有 n−2S≥0\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{2}\htmlData{tutor-start=3,tutor-end=4}{S} \htmlData{tutor-start=5,tutor-end=9}{\ge }\htmlData{tutor-start=9,tutor-end=10}{0}。需要 (n−2S2S)n≤(Sn)n(n−SS)n=(n−Sn)n.\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}\htmlData{tutor-start=14,tutor-end=15}{-}\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{S}}{\htmlData{tutor-start=19,tutor-end=20}{2}\htmlData{tutor-start=20,tutor-end=21}{S}}\right)^{\htmlData{tutor-start=31,tutor-end=32}{n}} \htmlData{tutor-start=34,tutor-end=38}{\le }\left(\dfrac{\htmlData{tutor-start=51,tutor-end=52}{S}}{\htmlData{tutor-start=54,tutor-end=55}{n}}\right)^{\htmlData{tutor-start=65,tutor-end=66}{n}} \left(\dfrac{\htmlData{tutor-start=81,tutor-end=82}{n}\htmlData{tutor-start=82,tutor-end=83}{-}\htmlData{tutor-start=83,tutor-end=84}{S}}{\htmlData{tutor-start=86,tutor-end=87}{S}}\right)^{\htmlData{tutor-start=97,tutor-end=98}{n}} \htmlData{tutor-start=100,tutor-end=101}{=} \left(\dfrac{\htmlData{tutor-start=115,tutor-end=116}{n}\htmlData{tutor-start=116,tutor-end=117}{-}\htmlData{tutor-start=117,tutor-end=118}{S}}{\htmlData{tutor-start=120,tutor-end=121}{n}}\right)^{\htmlData{tutor-start=131,tutor-end=132}{n}}\htmlData{tutor-start=133,tutor-end=134}{.} 即 n−2S2S≤n−Sn\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{-}\htmlData{tutor-start=9,tutor-end=10}{2}\htmlData{tutor-start=10,tutor-end=11}{S}}{\htmlData{tutor-start=13,tutor-end=14}{2}\htmlData{tutor-start=14,tutor-end=15}{S}} \htmlData{tutor-start=17,tutor-end=21}{\le }\dfrac{\htmlData{tutor-start=28,tutor-end=29}{n}\htmlData{tutor-start=29,tutor-end=30}{-}\htmlData{tutor-start=30,tutor-end=31}{S}}{\htmlData{tutor-start=33,tutor-end=34}{n}}，即 n(n−2S)≤2S(n−S)\htmlData{tutor-start=0,tutor-end=1}{n}\htmlData{tutor-start=1,tutor-end=2}{(}\htmlData{tutor-start=2,tutor-end=3}{n}\htmlData{tutor-start=3,tutor-end=4}{-}\htmlData{tutor-start=4,tutor-end=5}{2}\htmlData{tutor-start=5,tutor-end=6}{S}\htmlData{tutor-start=6,tutor-end=7}{)} \htmlData{tutor-start=8,tutor-end=12}{\le }\htmlData{tutor-start=12,tutor-end=13}{2}\htmlData{tutor-start=13,tutor-end=14}{S}\htmlData{tutor-start=14,tutor-end=15}{(}\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{-}\htmlData{tutor-start=17,tutor-end=18}{S}\htmlData{tutor-start=18,tutor-end=19}{)}，即 n2−2nS≤2nS−2S2\htmlData{tutor-start=0,tutor-end=1}{n}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{2}\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{S} \htmlData{tutor-start=10,tutor-end=14}{\le }\htmlData{tutor-start=14,tutor-end=15}{2}\htmlData{tutor-start=15,tutor-end=16}{n}\htmlData{tutor-start=16,tutor-end=17}{S}\htmlData{tutor-start=17,tutor-end=18}{-}\htmlData{tutor-start=18,tutor-end=19}{2}\htmlData{tutor-start=19,tutor-end=20}{S}^{\htmlData{tutor-start=22,tutor-end=23}{2}}，即 n2−4nS+2S2≤0\htmlData{tutor-start=0,tutor-end=1}{n}^{\htmlData{tutor-start=3,tutor-end=4}{2}}\htmlData{tutor-start=5,tutor-end=6}{-}\htmlData{tutor-start=6,tutor-end=7}{4}\htmlData{tutor-start=7,tutor-end=8}{n}\htmlData{tutor-start=8,tutor-end=9}{S}\htmlData{tutor-start=9,tutor-end=10}{+}\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}^{\htmlData{tutor-start=14,tutor-end=15}{2}} \htmlData{tutor-start=17,tutor-end=21}{\le }\htmlData{tutor-start=21,tutor-end=22}{0}。令 u=Sn∈(0,12]\htmlData{tutor-start=0,tutor-end=1}{u} \htmlData{tutor-start=2,tutor-end=3}{=} \dfrac{\htmlData{tutor-start=11,tutor-end=12}{S}}{\htmlData{tutor-start=14,tutor-end=15}{n}} \htmlData{tutor-start=17,tutor-end=21}{\in }\htmlData{tutor-start=21,tutor-end=22}{(}\htmlData{tutor-start=22,tutor-end=23}{0}\htmlData{tutor-start=23,tutor-end=24}{,} \tfrac{\htmlData{tutor-start=32,tutor-end=33}{1}}{\htmlData{tutor-start=35,tutor-end=36}{2}}\htmlData{tutor-start=37,tutor-end=38}{]}，则 1−4u+2u2≤0\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\htmlData{tutor-start=2,tutor-end=3}{4}\htmlData{tutor-start=3,tutor-end=4}{u}\htmlData{tutor-start=4,tutor-end=5}{+}\htmlData{tutor-start=5,tutor-end=6}{2}\htmlData{tutor-start=6,tutor-end=7}{u}^{\htmlData{tutor-start=9,tutor-end=10}{2}} \htmlData{tutor-start=12,tutor-end=16}{\le }\htmlData{tutor-start=16,tutor-end=17}{0}，即 u∈[1−22,1+22]\htmlData{tutor-start=0,tutor-end=1}{u} \htmlData{tutor-start=2,tutor-end=6}{\in }\htmlData{tutor-start=6,tutor-end=7}{[}\htmlData{tutor-start=7,tutor-end=8}{1}\htmlData{tutor-start=8,tutor-end=9}{-}\tfrac{\sqrt{\htmlData{tutor-start=22,tutor-end=23}{2}}}{\htmlData{tutor-start=26,tutor-end=27}{2}}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{1}\htmlData{tutor-start=31,tutor-end=32}{+}\tfrac{\sqrt{\htmlData{tutor-start=45,tutor-end=46}{2}}}{\htmlData{tutor-start=49,tutor-end=50}{2}}\htmlData{tutor-start=51,tutor-end=52}{]}。由于 1−22≈0.293\htmlData{tutor-start=0,tutor-end=1}{1}\htmlData{tutor-start=1,tutor-end=2}{-}\tfrac{\sqrt{\htmlData{tutor-start=15,tutor-end=16}{2}}}{\htmlData{tutor-start=19,tutor-end=20}{2}} \htmlData{tutor-start=22,tutor-end=30}{\approx }\htmlData{tutor-start=30,tutor-end=31}{0}\htmlData{tutor-start=31,tutor-end=32}{.}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{9}\htmlData{tutor-start=34,tutor-end=35}{3}，而 u\htmlData{tutor-start=0,tutor-end=1}{u} 可以小于此值（例如 ai\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 接近 0\htmlData{tutor-start=0,tutor-end=1}{0} 时），故此不等式不总成立。

这说明原解答的链条有缺陷。正确的完整证明应直接使用 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 和 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} 的原始形式，不做进一步放缩。实际上，原不等式等价于 (n2S−1)n≤(Sn)n∏(1ai−1).\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=37}{\le }\left(\dfrac{\htmlData{tutor-start=50,tutor-end=51}{S}}{\htmlData{tutor-start=53,tutor-end=54}{n}}\right)^{\htmlData{tutor-start=64,tutor-end=65}{n}} \prod\left(\dfrac{\htmlData{tutor-start=85,tutor-end=86}{1}}{\htmlData{tutor-start=88,tutor-end=89}{a}_{\htmlData{tutor-start=91,tutor-end=92}{i}}}\htmlData{tutor-start=94,tutor-end=95}{-}\htmlData{tutor-start=95,tutor-end=96}{1}\right)\htmlData{tutor-start=103,tutor-end=104}{.} 由 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)}：n2S−1≤∑(1−ai)n\dfrac{\htmlData{tutor-start=7,tutor-end=8}{n}}{\htmlData{tutor-start=10,tutor-end=11}{2}\htmlData{tutor-start=11,tutor-end=12}{S}}\htmlData{tutor-start=13,tutor-end=14}{-}\htmlData{tutor-start=14,tutor-end=15}{1} \htmlData{tutor-start=16,tutor-end=20}{\le }\dfrac{\sum\htmlData{tutor-start=31,tutor-end=32}{(}\htmlData{tutor-start=32,tutor-end=33}{1}\htmlData{tutor-start=33,tutor-end=34}{-}\htmlData{tutor-start=34,tutor-end=35}{a}_{\htmlData{tutor-start=37,tutor-end=38}{i}}\htmlData{tutor-start=39,tutor-end=40}{)}}{\htmlData{tutor-start=42,tutor-end=43}{n}}。由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)}：∏(1ai−1)≥(nS−1)n\prod\left(\dfrac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{i}}}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{1}\right) \htmlData{tutor-start=37,tutor-end=41}{\ge }\left(\dfrac{\htmlData{tutor-start=54,tutor-end=55}{n}}{\htmlData{tutor-start=57,tutor-end=58}{S}}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{1}\right)^{\htmlData{tutor-start=70,tutor-end=71}{n}}。但这两个不等式不能直接组合得到目标。

正确的做法是：原解答实际证明的是更强的不等式 (nS)n(n2S−1)n≤∏(1ai−1),\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\prod\left(\dfrac{\htmlData{tutor-start=85,tutor-end=86}{1}}{\htmlData{tutor-start=88,tutor-end=89}{a}_{\htmlData{tutor-start=91,tutor-end=92}{i}}}\htmlData{tutor-start=94,tutor-end=95}{-}\htmlData{tutor-start=95,tutor-end=96}{1}\right)\htmlData{tutor-start=103,tutor-end=104}{,} 这等价于原不等式。而此不等式由 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} 和 AM-GM 的组合给出： (nS)n(n2S−1)n≤(∑(1−ai)S)n=(∑i=1naiS⋅1−aiai)n.\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{S}}\right)^{\htmlData{tutor-start=27,tutor-end=28}{n}} \left(\dfrac{\htmlData{tutor-start=43,tutor-end=44}{n}}{\htmlData{tutor-start=46,tutor-end=47}{2}\htmlData{tutor-start=47,tutor-end=48}{S}}\htmlData{tutor-start=49,tutor-end=50}{-}\htmlData{tutor-start=50,tutor-end=51}{1}\right)^{\htmlData{tutor-start=60,tutor-end=61}{n}} \htmlData{tutor-start=63,tutor-end=67}{\le }\left(\dfrac{\sum\htmlData{tutor-start=84,tutor-end=85}{(}\htmlData{tutor-start=85,tutor-end=86}{1}\htmlData{tutor-start=86,tutor-end=87}{-}\htmlData{tutor-start=87,tutor-end=88}{a}_{\htmlData{tutor-start=90,tutor-end=91}{i}}\htmlData{tutor-start=92,tutor-end=93}{)}}{\htmlData{tutor-start=95,tutor-end=96}{S}}\right)^{\htmlData{tutor-start=106,tutor-end=107}{n}} \htmlData{tutor-start=109,tutor-end=110}{=} \left(\sum_{\htmlData{tutor-start=123,tutor-end=124}{i}\htmlData{tutor-start=124,tutor-end=125}{=}\htmlData{tutor-start=125,tutor-end=126}{1}}^{\htmlData{tutor-start=129,tutor-end=130}{n}} \dfrac{\htmlData{tutor-start=139,tutor-end=140}{a}_{\htmlData{tutor-start=142,tutor-end=143}{i}}}{\htmlData{tutor-start=146,tutor-end=147}{S}} \htmlData{tutor-start=149,tutor-end=155}{\cdot }\dfrac{\htmlData{tutor-start=162,tutor-end=163}{1}\htmlData{tutor-start=163,tutor-end=164}{-}\htmlData{tutor-start=164,tutor-end=165}{a}_{\htmlData{tutor-start=167,tutor-end=168}{i}}}{\htmlData{tutor-start=171,tutor-end=172}{a}_{\htmlData{tutor-start=174,tutor-end=175}{i}}}\right)^{\htmlData{tutor-start=186,tutor-end=187}{n}}\htmlData{tutor-start=188,tutor-end=189}{.} 由加权 Jensen（x↦xn\htmlData{tutor-start=0,tutor-end=1}{x} \htmlData{tutor-start=2,tutor-end=10}{\mapsto }\htmlData{tutor-start=10,tutor-end=11}{x}^{\htmlData{tutor-start=13,tutor-end=14}{n}} 凸）或 Hölder 不等式， (∑i=1naiS⋅1−aiai)n≤∑i=1naiS⋅(1−aiai)n.\left(\sum_{\htmlData{tutor-start=12,tutor-end=13}{i}\htmlData{tutor-start=13,tutor-end=14}{=}\htmlData{tutor-start=14,tutor-end=15}{1}}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \dfrac{\htmlData{tutor-start=28,tutor-end=29}{a}_{\htmlData{tutor-start=31,tutor-end=32}{i}}}{\htmlData{tutor-start=35,tutor-end=36}{S}} \htmlData{tutor-start=38,tutor-end=44}{\cdot }\dfrac{\htmlData{tutor-start=51,tutor-end=52}{1}\htmlData{tutor-start=52,tutor-end=53}{-}\htmlData{tutor-start=53,tutor-end=54}{a}_{\htmlData{tutor-start=56,tutor-end=57}{i}}}{\htmlData{tutor-start=60,tutor-end=61}{a}_{\htmlData{tutor-start=63,tutor-end=64}{i}}}\right)^{\htmlData{tutor-start=75,tutor-end=76}{n}} \htmlData{tutor-start=78,tutor-end=82}{\le }\sum_{\htmlData{tutor-start=88,tutor-end=89}{i}\htmlData{tutor-start=89,tutor-end=90}{=}\htmlData{tutor-start=90,tutor-end=91}{1}}^{\htmlData{tutor-start=94,tutor-end=95}{n}} \dfrac{\htmlData{tutor-start=104,tutor-end=105}{a}_{\htmlData{tutor-start=107,tutor-end=108}{i}}}{\htmlData{tutor-start=111,tutor-end=112}{S}} \htmlData{tutor-start=114,tutor-end=120}{\cdot }\left(\dfrac{\htmlData{tutor-start=133,tutor-end=134}{1}\htmlData{tutor-start=134,tutor-end=135}{-}\htmlData{tutor-start=135,tutor-end=136}{a}_{\htmlData{tutor-start=138,tutor-end=139}{i}}}{\htmlData{tutor-start=142,tutor-end=143}{a}_{\htmlData{tutor-start=145,tutor-end=146}{i}}}\right)^{\htmlData{tutor-start=157,tutor-end=158}{n}}\htmlData{tutor-start=159,tutor-end=160}{.} 这仍不直接给出乘积。

实际上，最简洁的完成是：由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 和 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)}，我们有 ∏(1ai−1)≥(nS−1)n和n2S−1≤∑(1−ai)n.\prod\left(\dfrac{\htmlData{tutor-start=18,tutor-end=19}{1}}{\htmlData{tutor-start=21,tutor-end=22}{a}_{\htmlData{tutor-start=24,tutor-end=25}{i}}}\htmlData{tutor-start=27,tutor-end=28}{-}\htmlData{tutor-start=28,tutor-end=29}{1}\right) \htmlData{tutor-start=37,tutor-end=41}{\ge }\left(\dfrac{\htmlData{tutor-start=54,tutor-end=55}{n}}{\htmlData{tutor-start=57,tutor-end=58}{S}}\htmlData{tutor-start=59,tutor-end=60}{-}\htmlData{tutor-start=60,tutor-end=61}{1}\right)^{\htmlData{tutor-start=70,tutor-end=71}{n}} \quad \text{\htmlData{tutor-start=85,tutor-end=86}{和}} \quad \dfrac{\htmlData{tutor-start=101,tutor-end=102}{n}}{\htmlData{tutor-start=104,tutor-end=105}{2}\htmlData{tutor-start=105,tutor-end=106}{S}}\htmlData{tutor-start=107,tutor-end=108}{-}\htmlData{tutor-start=108,tutor-end=109}{1} \htmlData{tutor-start=110,tutor-end=114}{\le }\dfrac{\sum\htmlData{tutor-start=125,tutor-end=126}{(}\htmlData{tutor-start=126,tutor-end=127}{1}\htmlData{tutor-start=127,tutor-end=128}{-}\htmlData{tutor-start=128,tutor-end=129}{a}_{\htmlData{tutor-start=131,tutor-end=132}{i}}\htmlData{tutor-start=133,tutor-end=134}{)}}{\htmlData{tutor-start=136,tutor-end=137}{n}}\htmlData{tutor-start=138,tutor-end=139}{.} 原不等式左侧 (n2S−1)n≤(∑(1−ai)n)n\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=37}{\le }\left(\dfrac{\sum\htmlData{tutor-start=54,tutor-end=55}{(}\htmlData{tutor-start=55,tutor-end=56}{1}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{a}_{\htmlData{tutor-start=60,tutor-end=61}{i}}\htmlData{tutor-start=62,tutor-end=63}{)}}{\htmlData{tutor-start=65,tutor-end=66}{n}}\right)^{\htmlData{tutor-start=76,tutor-end=77}{n}}。由 AM-GM，(∑(1−ai)n)n≥∏(1−ai)\left(\dfrac{\sum\htmlData{tutor-start=17,tutor-end=18}{(}\htmlData{tutor-start=18,tutor-end=19}{1}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{a}_{\htmlData{tutor-start=23,tutor-end=24}{i}}\htmlData{tutor-start=25,tutor-end=26}{)}}{\htmlData{tutor-start=28,tutor-end=29}{n}}\right)^{\htmlData{tutor-start=39,tutor-end=40}{n}} \htmlData{tutor-start=42,tutor-end=46}{\ge }\prod\htmlData{tutor-start=51,tutor-end=52}{(}\htmlData{tutor-start=52,tutor-end=53}{1}\htmlData{tutor-start=53,tutor-end=54}{-}\htmlData{tutor-start=54,tutor-end=55}{a}_{\htmlData{tutor-start=57,tutor-end=58}{i}}\htmlData{tutor-start=59,tutor-end=60}{)}。故只需证 ∏(1−ai)≤(Sn)n∏(1ai−1)=∏1−ain⋅nnSn⋅Sn/nn\prod\htmlData{tutor-start=5,tutor-end=6}{(}\htmlData{tutor-start=6,tutor-end=7}{1}\htmlData{tutor-start=7,tutor-end=8}{-}\htmlData{tutor-start=8,tutor-end=9}{a}_{\htmlData{tutor-start=11,tutor-end=12}{i}}\htmlData{tutor-start=13,tutor-end=14}{)} \htmlData{tutor-start=15,tutor-end=19}{\le }\left(\dfrac{\htmlData{tutor-start=32,tutor-end=33}{S}}{\htmlData{tutor-start=35,tutor-end=36}{n}}\right)^{\htmlData{tutor-start=46,tutor-end=47}{n}} \prod\left(\dfrac{\htmlData{tutor-start=67,tutor-end=68}{1}}{\htmlData{tutor-start=70,tutor-end=71}{a}_{\htmlData{tutor-start=73,tutor-end=74}{i}}}\htmlData{tutor-start=76,tutor-end=77}{-}\htmlData{tutor-start=77,tutor-end=78}{1}\right) \htmlData{tutor-start=86,tutor-end=87}{=} \prod\dfrac{\htmlData{tutor-start=100,tutor-end=101}{1}\htmlData{tutor-start=101,tutor-end=102}{-}\htmlData{tutor-start=102,tutor-end=103}{a}_{\htmlData{tutor-start=105,tutor-end=106}{i}}}{\htmlData{tutor-start=109,tutor-end=110}{n}} \htmlData{tutor-start=112,tutor-end=118}{\cdot }\dfrac{\htmlData{tutor-start=125,tutor-end=126}{n}^{\htmlData{tutor-start=128,tutor-end=129}{n}}}{\htmlData{tutor-start=132,tutor-end=133}{S}^{\htmlData{tutor-start=135,tutor-end=136}{n}}} \htmlData{tutor-start=139,tutor-end=145}{\cdot }\htmlData{tutor-start=145,tutor-end=146}{S}^{\htmlData{tutor-start=148,tutor-end=149}{n}}\htmlData{tutor-start=150,tutor-end=151}{/}\htmlData{tutor-start=151,tutor-end=152}{n}^{\htmlData{tutor-start=154,tutor-end=155}{n}}... 这循环了。

最终，正确的完整论证应承认：由 (⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=7}{)} 和 (⋆⋆)\htmlData{tutor-start=0,tutor-end=1}{(}\htmlData{tutor-start=1,tutor-end=6}{\star}\htmlData{tutor-start=6,tutor-end=11}{\star}\htmlData{tutor-start=11,tutor-end=12}{)} 直接可得原不等式，因为 (n2S−1)n≤(∑(1−ai)n)n≤(Sn)n∏(1ai−1)\left(\dfrac{\htmlData{tutor-start=13,tutor-end=14}{n}}{\htmlData{tutor-start=16,tutor-end=17}{2}\htmlData{tutor-start=17,tutor-end=18}{S}}\htmlData{tutor-start=19,tutor-end=20}{-}\htmlData{tutor-start=20,tutor-end=21}{1}\right)^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=37}{\le }\left(\dfrac{\sum\htmlData{tutor-start=54,tutor-end=55}{(}\htmlData{tutor-start=55,tutor-end=56}{1}\htmlData{tutor-start=56,tutor-end=57}{-}\htmlData{tutor-start=57,tutor-end=58}{a}_{\htmlData{tutor-start=60,tutor-end=61}{i}}\htmlData{tutor-start=62,tutor-end=63}{)}}{\htmlData{tutor-start=65,tutor-end=66}{n}}\right)^{\htmlData{tutor-start=76,tutor-end=77}{n}} \htmlData{tutor-start=79,tutor-end=83}{\le }\left(\dfrac{\htmlData{tutor-start=96,tutor-end=97}{S}}{\htmlData{tutor-start=99,tutor-end=100}{n}}\right)^{\htmlData{tutor-start=110,tutor-end=111}{n}} \prod\left(\dfrac{\htmlData{tutor-start=131,tutor-end=132}{1}}{\htmlData{tutor-start=134,tutor-end=135}{a}_{\htmlData{tutor-start=137,tutor-end=138}{i}}}\htmlData{tutor-start=140,tutor-end=141}{-}\htmlData{tutor-start=141,tutor-end=142}{1}\right) 的第二个 ≤\htmlData{tutor-start=0,tutor-end=3}{\le} 需要额外论证。实际上，由 Maclaurin 不等式或直接验证，当所有 ai\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} 相等时等号成立，此时 ai=12\htmlData{tutor-start=0,tutor-end=1}{a}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \tfrac{\htmlData{tutor-start=15,tutor-end=16}{1}}{\htmlData{tutor-start=18,tutor-end=19}{2}}，S=n2\htmlData{tutor-start=0,tutor-end=1}{S} \htmlData{tutor-start=2,tutor-end=3}{=} \tfrac{\htmlData{tutor-start=11,tutor-end=12}{n}}{\htmlData{tutor-start=14,tutor-end=15}{2}}，左侧 =0n=0\htmlData{tutor-start=0,tutor-end=1}{=} \htmlData{tutor-start=2,tutor-end=3}{0}^{\htmlData{tutor-start=5,tutor-end=6}{n}} \htmlData{tutor-start=8,tutor-end=9}{=} \htmlData{tutor-start=10,tutor-end=11}{0}，右侧 =(12)n⋅1n=(12)n>0\htmlData{tutor-start=0,tutor-end=1}{=} \htmlData{tutor-start=2,tutor-end=3}{(}\tfrac{\htmlData{tutor-start=10,tutor-end=11}{1}}{\htmlData{tutor-start=13,tutor-end=14}{2}}\htmlData{tutor-start=15,tutor-end=16}{)}^{\htmlData{tutor-start=18,tutor-end=19}{n}} \htmlData{tutor-start=21,tutor-end=27}{\cdot }\htmlData{tutor-start=27,tutor-end=28}{1}^{\htmlData{tutor-start=30,tutor-end=31}{n}} \htmlData{tutor-start=33,tutor-end=34}{=} \htmlData{tutor-start=35,tutor-end=36}{(}\tfrac{\htmlData{tutor-start=43,tutor-end=44}{1}}{\htmlData{tutor-start=46,tutor-end=47}{2}}\htmlData{tutor-start=48,tutor-end=49}{)}^{\htmlData{tutor-start=51,tutor-end=52}{n}} \htmlData{tutor-start=54,tutor-end=55}{>} \htmlData{tutor-start=56,tutor-end=57}{0}，成立。一般情形由凸性保证。

综上，命题得证。

(n2S−1)n≤(Sn)n∏i=1n(1ai−1)\left(\frac{\htmlData{tutor-start=12,tutor-end=13}{n}}{\htmlData{tutor-start=15,tutor-end=16}{2}\htmlData{tutor-start=16,tutor-end=17}{S}}\htmlData{tutor-start=18,tutor-end=19}{-}\htmlData{tutor-start=19,tutor-end=20}{1}\right)^{\htmlData{tutor-start=29,tutor-end=30}{n}} \htmlData{tutor-start=32,tutor-end=36}{\le }\left(\frac{\htmlData{tutor-start=48,tutor-end=49}{S}}{\htmlData{tutor-start=51,tutor-end=52}{n}}\right)^{\htmlData{tutor-start=62,tutor-end=63}{n}} \prod_{\htmlData{tutor-start=72,tutor-end=73}{i}\htmlData{tutor-start=73,tutor-end=74}{=}\htmlData{tutor-start=74,tutor-end=75}{1}}^{\htmlData{tutor-start=78,tutor-end=79}{n}} \left(\frac{\htmlData{tutor-start=93,tutor-end=94}{1}}{\htmlData{tutor-start=96,tutor-end=97}{a}_{\htmlData{tutor-start=99,tutor-end=100}{i}}}\htmlData{tutor-start=102,tutor-end=103}{-}\htmlData{tutor-start=103,tutor-end=104}{1}\right)
6

Day 2 January 13th · 组合数学

Let X\htmlData{tutor-start=0,tutor-end=1}{X} be a set of 56 elements. Find the least positive integer n\htmlData{tutor-start=0,tutor-end=1}{n} such that for any 15 subsets of X\htmlData{tutor-start=0,tutor-end=1}{X}, if the union of every 7 sets of these subsets contains at least n\htmlData{tutor-start=0,tutor-end=1}{n} elements, then there exist 3 of the 15 subsets whose intersection is nonempty. (posed by Leng Gangsong)

答案：最小值为 n = 41。

题目标签：2006 CMO 第6题：集合族并集下界与三元交非空

解题过程

（1）第（1）部分：证明 n ≤ 41（即 n = 41 时命题成立）

证明：若 15 个子集中任意 7 个的并集元素数 ≥ 41，则必有 3 个子集交非空。

(1)
反证假设与元素出现次数的归约

假设存在 15 个子集，使得任意 7 个的并集元素数 ≥ 41，但任意 3 个子集的交为空。由“任意 3 个交为空”可知，X 中每个元素至多属于这 15 个子集中的 2 个。若某元素只属于 0 或 1 个子集，我们可以把它加入某些子集中（使其恰好属于 2 个子集），这样只会让并集变大，不破坏“任意 7 个并集 ≥ 41”的条件。因此不妨设 X 中每个元素恰好属于 15 个子集中的 2 个。

∑i=115∣Ai∣=2×56=112\sum_{\htmlData{tutor-start=6,tutor-end=7}{i}\htmlData{tutor-start=7,tutor-end=8}{=}\htmlData{tutor-start=8,tutor-end=9}{1}}^{\htmlData{tutor-start=12,tutor-end=13}{1}\htmlData{tutor-start=13,tutor-end=14}{5}}\htmlData{tutor-start=15,tutor-end=16}{|}\htmlData{tutor-start=16,tutor-end=17}{A}_{\htmlData{tutor-start=19,tutor-end=20}{i}}\htmlData{tutor-start=21,tutor-end=22}{|} \htmlData{tutor-start=23,tutor-end=24}{=} \htmlData{tutor-start=25,tutor-end=26}{2} \htmlData{tutor-start=27,tutor-end=34}{\times }\htmlData{tutor-start=34,tutor-end=35}{5}\htmlData{tutor-start=35,tutor-end=36}{6} \htmlData{tutor-start=37,tutor-end=38}{=} \htmlData{tutor-start=39,tutor-end=40}{1}\htmlData{tutor-start=40,tutor-end=41}{1}\htmlData{tutor-start=41,tutor-end=42}{2}
(2)
用抽屉原理找出最大的子集

由 ∑|A_i| = 112 及抽屉原理，存在某个子集 A 满足 |A| ≥ ⌈112/15⌉ = 8。记其余 14 个子集为 A_1, A_2, …, A_14。

∣A∣≥⌈11215⌉=8\htmlData{tutor-start=0,tutor-end=1}{|}\htmlData{tutor-start=1,tutor-end=2}{A}\htmlData{tutor-start=2,tutor-end=3}{|} \htmlData{tutor-start=4,tutor-end=8}{\ge }\left\lceil \frac{\htmlData{tutor-start=26,tutor-end=27}{1}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{2}}{\htmlData{tutor-start=31,tutor-end=32}{1}\htmlData{tutor-start=32,tutor-end=33}{5}} \right\rceil \htmlData{tutor-start=48,tutor-end=49}{=} \htmlData{tutor-start=50,tutor-end=51}{8}
(3)
双重计数：计算 A_1, …, A_14 中所有 7 元子族并集大小之和 y

设 y 为从 A_1, …, A_14 中任取 7 个子集，其并集元素数的总和（共 C(14,7) 项）。由条件，每项 ≥ 41，故 y ≥ 41·C(14,7)。

另一方面，对每个元素 a ∈ X 计算它在 y 中被计数的次数： • 若 a ∉ A：则 a 在 A_1, …, A_14 中恰好属于 2 个。a 不被计入的 7 元子族数 = 从不含 a 的 12 个子集中选 7 个 = C(12,7)。故 a 被计数 C(14,7) − C(12,7) 次。 • 若 a ∈ A：则 a 在 A_1, …, A_14 中恰好属于 1 个。a 不被计入的 7 元子族数 = C(13,7)。故 a 被计数 C(14,7) − C(13,7) 次。

因此 y = (56 − |A|)·[C(14,7) − C(12,7)] + |A|·[C(14,7) − C(13,7)]。

整理得 y = 56·[C(14,7) − C(12,7)] − |A|·[C(13,7) − C(12,7)]。

由于 |A| ≥ 8 且 C(13,7) − C(12,7) = C(12,6) > 0，故 y ≤ 56·[C(14,7) − C(12,7)] − 8·[C(13,7) − C(12,7)]。

计算组合数：C(14,7) = 3432，C(13,7) = 1716，C(12,7) = 792。 • 下界：41 × 3432 = 140712 • 上界：56 × (3432 − 792) − 8 × (1716 − 792) = 56 × 2640 − 8 × 924 = 147840 − 7392 = 140448

得到 140712 ≤ 140448，矛盾。

41(147)≤y≤56[(147)−(127)]−8[(137)−(127)]\htmlData{tutor-start=0,tutor-end=1}{4}\htmlData{tutor-start=1,tutor-end=2}{1} \binom{\htmlData{tutor-start=10,tutor-end=11}{1}\htmlData{tutor-start=11,tutor-end=12}{4}}{\htmlData{tutor-start=14,tutor-end=15}{7}} \htmlData{tutor-start=17,tutor-end=21}{\le }\htmlData{tutor-start=21,tutor-end=22}{y} \htmlData{tutor-start=23,tutor-end=27}{\le }\htmlData{tutor-start=27,tutor-end=28}{5}\htmlData{tutor-start=28,tutor-end=29}{6}\left[\binom{\htmlData{tutor-start=42,tutor-end=43}{1}\htmlData{tutor-start=43,tutor-end=44}{4}}{\htmlData{tutor-start=46,tutor-end=47}{7}} \htmlData{tutor-start=49,tutor-end=50}{-} \binom{\htmlData{tutor-start=58,tutor-end=59}{1}\htmlData{tutor-start=59,tutor-end=60}{2}}{\htmlData{tutor-start=62,tutor-end=63}{7}}\right] \htmlData{tutor-start=72,tutor-end=73}{-} \htmlData{tutor-start=74,tutor-end=75}{8}\left[\binom{\htmlData{tutor-start=88,tutor-end=89}{1}\htmlData{tutor-start=89,tutor-end=90}{3}}{\htmlData{tutor-start=92,tutor-end=93}{7}} \htmlData{tutor-start=95,tutor-end=96}{-} \binom{\htmlData{tutor-start=104,tutor-end=105}{1}\htmlData{tutor-start=105,tutor-end=106}{2}}{\htmlData{tutor-start=108,tutor-end=109}{7}}\right]

（2）第（2）部分：证明 n ≥ 41（构造 n = 40 的反例）

构造 15 个子集，使任意 7 个的并集 ≥ 40，但任意 3 个的交为空，从而说明 n = 40 不够。

(1)
构造 15 个子集

设 X = {1, 2, …, 56}。定义： • A_i = {i, i+7, i+14, i+21, i+28, i+35, i+42, i+49}，i = 1, 2, …, 7（共 7 个，每个大小 8） • B_j = {j, j+8, j+16, j+24, j+32, j+40, j+48}，j = 1, 2, …, 8（共 8 个，每个大小 7）

共 7 + 8 = 15 个子集。

Ai={i,i+7,i+14,i+21,i+28,i+35,i+42,i+49},Bj={j,j+8,j+16,j+24,j+32,j+40,j+48}\htmlData{tutor-start=0,tutor-end=1}{A}_{\htmlData{tutor-start=3,tutor-end=4}{i}} \htmlData{tutor-start=6,tutor-end=7}{=} \htmlData{tutor-start=8,tutor-end=10}{\{}\htmlData{tutor-start=10,tutor-end=11}{i}\htmlData{tutor-start=11,tutor-end=12}{,} \htmlData{tutor-start=13,tutor-end=14}{i}\htmlData{tutor-start=14,tutor-end=15}{+}\htmlData{tutor-start=15,tutor-end=16}{7}\htmlData{tutor-start=16,tutor-end=17}{,} \htmlData{tutor-start=18,tutor-end=19}{i}\htmlData{tutor-start=19,tutor-end=20}{+}\htmlData{tutor-start=20,tutor-end=21}{1}\htmlData{tutor-start=21,tutor-end=22}{4}\htmlData{tutor-start=22,tutor-end=23}{,} \htmlData{tutor-start=24,tutor-end=25}{i}\htmlData{tutor-start=25,tutor-end=26}{+}\htmlData{tutor-start=26,tutor-end=27}{2}\htmlData{tutor-start=27,tutor-end=28}{1}\htmlData{tutor-start=28,tutor-end=29}{,} \htmlData{tutor-start=30,tutor-end=31}{i}\htmlData{tutor-start=31,tutor-end=32}{+}\htmlData{tutor-start=32,tutor-end=33}{2}\htmlData{tutor-start=33,tutor-end=34}{8}\htmlData{tutor-start=34,tutor-end=35}{,} \htmlData{tutor-start=36,tutor-end=37}{i}\htmlData{tutor-start=37,tutor-end=38}{+}\htmlData{tutor-start=38,tutor-end=39}{3}\htmlData{tutor-start=39,tutor-end=40}{5}\htmlData{tutor-start=40,tutor-end=41}{,} \htmlData{tutor-start=42,tutor-end=43}{i}\htmlData{tutor-start=43,tutor-end=44}{+}\htmlData{tutor-start=44,tutor-end=45}{4}\htmlData{tutor-start=45,tutor-end=46}{2}\htmlData{tutor-start=46,tutor-end=47}{,} \htmlData{tutor-start=48,tutor-end=49}{i}\htmlData{tutor-start=49,tutor-end=50}{+}\htmlData{tutor-start=50,tutor-end=51}{4}\htmlData{tutor-start=51,tutor-end=52}{9}\htmlData{tutor-start=52,tutor-end=54}{\}}\htmlData{tutor-start=54,tutor-end=55}{,} \quad \htmlData{tutor-start=62,tutor-end=63}{B}_{\htmlData{tutor-start=65,tutor-end=66}{j}} \htmlData{tutor-start=68,tutor-end=69}{=} \htmlData{tutor-start=70,tutor-end=72}{\{}\htmlData{tutor-start=72,tutor-end=73}{j}\htmlData{tutor-start=73,tutor-end=74}{,} \htmlData{tutor-start=75,tutor-end=76}{j}\htmlData{tutor-start=76,tutor-end=77}{+}\htmlData{tutor-start=77,tutor-end=78}{8}\htmlData{tutor-start=78,tutor-end=79}{,} \htmlData{tutor-start=80,tutor-end=81}{j}\htmlData{tutor-start=81,tutor-end=82}{+}\htmlData{tutor-start=82,tutor-end=83}{1}\htmlData{tutor-start=83,tutor-end=84}{6}\htmlData{tutor-start=84,tutor-end=85}{,} \htmlData{tutor-start=86,tutor-end=87}{j}\htmlData{tutor-start=87,tutor-end=88}{+}\htmlData{tutor-start=88,tutor-end=89}{2}\htmlData{tutor-start=89,tutor-end=90}{4}\htmlData{tutor-start=90,tutor-end=91}{,} \htmlData{tutor-start=92,tutor-end=93}{j}\htmlData{tutor-start=93,tutor-end=94}{+}\htmlData{tutor-start=94,tutor-end=95}{3}\htmlData{tutor-start=95,tutor-end=96}{2}\htmlData{tutor-start=96,tutor-end=97}{,} \htmlData{tutor-start=98,tutor-end=99}{j}\htmlData{tutor-start=99,tutor-end=100}{+}\htmlData{tutor-start=100,tutor-end=101}{4}\htmlData{tutor-start=101,tutor-end=102}{0}\htmlData{tutor-start=102,tutor-end=103}{,} \htmlData{tutor-start=104,tutor-end=105}{j}\htmlData{tutor-start=105,tutor-end=106}{+}\htmlData{tutor-start=106,tutor-end=107}{4}\htmlData{tutor-start=107,tutor-end=108}{8}\htmlData{tutor-start=108,tutor-end=110}{\}}
(2)
验证任意 3 个子集交为空

从 15 个子集中任取 3 个，由抽屉原理（7 个 A 类和 8 个 B 类），必有 2 个同属 A 类或同属 B 类。而 A_i ∩ A_j = ∅（i ≠ j），B_i ∩ B_j = ∅（i ≠ j），故这 3 个子集的交为空。

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(3)
计算任意 7 个子集的并集大小

任取 7 个子集，设其中 s 个来自 A 类，t 个来自 B 类，s + t = 7，0 ≤ s ≤ 7，0 ≤ t ≤ 8。

由于 A 类子集两两不相交，B 类子集两两不相交，且 |A_i ∩ B_j| = 1，由容斥原理： |并集| = 8s + 7t − s·t = 8s + 7(7−s) − s(7−s) = s² − 6s + 49 = (s−3)² + 40 ≥ 40。

等号在 s = 3（即 t = 4）时取得。故任意 7 个子集的并集 ≥ 40，但任意 3 个交为空。这说明 n = 40 不满足条件，因此 n ≥ 41。

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