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kuing
Post time 2015-3-5 15:16
应该是道老题，不过一时也没找到贴子，还是自己写写吧，反正均值就行了。
令 $x_i=y_i^n$, $y_i>0$，则也有 $y_1y_2\cdots y_n=1$，由均值
\begin{align*}
\frac1{n-1+x_i}&=\frac1{n-1}\left( 1-\frac{x_i}{n-1+x_i} \right) \\
& =\frac1{n-1}\left( 1-\frac{y_i^n}{(n-1)y_1y_2\cdots y_n+y_i^n} \right) \\
& =\frac1{n-1}\left( 1-\frac{y_i^{n-1}}{(n-1)y_1y_2\cdots y_{i-1}y_{i+1}\cdots y_n+y_i^{n-1}} \right) \\
& \leqslant \frac1{n-1}\left( 1-\frac{y_i^{n-1}}{y_1^{n-1}+y_2^{n-1}+\cdots +y_n^{n-1}} \right),
\end{align*}
所以
\[\sum_{i=1}^n\frac1{n-1+x_i}\leqslant \frac1{n-1}\left( n-\sum_{i=1}^n\frac{y_i^{n-1}}{y_1^{n-1}+y_2^{n-1}+\cdots +y_n^{n-1}} \right)=1.
\] |
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其妙
Post time 2015-3-5 19:39
