给定正实数 $a_1 , a_2 , \ldots , a_n$．证明：存在正实数 $x_1 , x_2 , \ldots, x_n$，满足 $\displaystyle \sum\limits_{i=1}^{n} x_{i}=1$，且对任何满足 $\displaystyle \sum\limits_{i=1}^{n} y_{i}=1$ 的正实数 $y_{1}, y_{2}, \ldots, y_{n}$，均有 $\displaystyle
\sum\limits_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+y_{i}} \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i}
$．
\sum\limits_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+y_{i}} \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i}
$．
【难度】
【出处】
2013第12届CGMO试题
【标注】
【答案】
略
【解析】
令 $\displaystyle x_{i}=\dfrac{a_{i}}{\sum\limits_{i=1}^{n} a_{i}}$，则 $\displaystyle \sum\limits_{i=1}^{n} x_{i}=1$．将 $x_i$ 代入不等式，得
$\displaystyle
\sum\limits_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+y_{i}}=\sum_{i=1}^{n} a_{i} \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}}
$
对任何正实数 $y_{1}, y_{2}, \cdots, y_{n}$，$\displaystyle \sum\limits_{i=1}^{n} y_{i}=1$，由柯西不等式可知
$\begin{aligned} 2 \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}} &=\sum_{i=1}^{n}\left(x_{i}+y_{i}\right) \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}} \\ & \geqslant\left(\sum_{i=1}^{n} x_{i}\right)^{2}=1 \end{aligned}$
于是，
$\displaystyle
\sum\limits_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+y_{i}}=\sum_{i=1}^{n} a_{i} \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}} \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i}
$
证毕．
$\displaystyle
\sum\limits_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+y_{i}}=\sum_{i=1}^{n} a_{i} \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}}
$
对任何正实数 $y_{1}, y_{2}, \cdots, y_{n}$，$\displaystyle \sum\limits_{i=1}^{n} y_{i}=1$，由柯西不等式可知
$\begin{aligned} 2 \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}} &=\sum_{i=1}^{n}\left(x_{i}+y_{i}\right) \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}} \\ & \geqslant\left(\sum_{i=1}^{n} x_{i}\right)^{2}=1 \end{aligned}$
于是，
$\displaystyle
\sum\limits_{i=1}^{n} \frac{a_{i} x_{i}}{x_{i}+y_{i}}=\sum_{i=1}^{n} a_{i} \sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+y_{i}} \geqslant \frac{1}{2} \sum_{i=1}^{n} a_{i}
$
证毕．
答案
解析
备注
