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[不等式] 转发一个不等式,5-10

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realnumber Posted 2016-5-10 09:13 |Read mode
Last edited by realnumber 2016-5-11 15:10 QQ图片20160510091224---.png
解答在三楼。2楼是另外一题

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 Author| realnumber Posted 2016-5-10 10:35
Last edited by realnumber 2016-5-11 12:53顶下,以前一个也放这里
QQ截图20160510103421666.jpg

由cauchy不等式
\[(\frac{1}{1+x_1}+\frac{1}{1+x_1+x_2}+\cdots +\frac{1}{1+x_1+\cdots +x_n})^2\le (\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n})(\frac{x_1}{(1+x_1)^2}+\frac{x_2}{(1+x_1+x_2)^2}+\cdots +\frac{x_n}{(1+x_1+\cdots +x_n)2})\]

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 Author| realnumber Posted 2016-5-10 14:21
哎,只需要证明这个就够了
\[\frac{i}{1+a_1+a_2+\cdots +a_i}\le \frac{1}{2}\sqrt{\frac{1}{a_1}+\cdots +\frac{1}{a_n}} \]
以下证明$n=i$时
\[4n^2\le (\frac{1}{a_1}+\cdots +\frac{1}{a_n})(1+a_1+\cdots +a_n)^2\]
\[4n^2\le (\frac{1}{a_1}+\cdots +\frac{1}{a_n}+n^2)(1+a_1+\cdots +a_n)\]
只需要证明
\[4n^2\le n^2+n^2+\frac{1}{a_1}+\cdots +\frac{1}{a_n}+n^2(a_1+\cdots +a_n)\]
完.

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