设 $x_{1}, x_{2}, \cdots, x_{n}(n \geqslant 2)$ 都是整数且 $\sum_{i=1}^{n} x_{i}=1$，求证 $\sum_{i=1}^{n} \dfrac{x_{i}}{\sqrt{1-x_{i}}} \geqslant \dfrac{\sum_{i=1}^{n} \sqrt{x_{i}}}{\sqrt{n-1}}$
【难度】
【出处】
1989第4届CMO试题
【标注】
【答案】
略
【解析】
$\begin{aligned} \sum_{i=1}^{n} \frac{x_{i}}{\sqrt{1-x_{i}}}=& \sum_{i=1}^{n} \frac{1}{\sqrt{1-x_{i}}}-\sum_{i=1}^{n} \frac{1-x_{i}}{\sqrt{1-x_{i}}}= \sum_{i=1}^{n} \frac{1}{\sqrt{1-x_{i}}}-\sum_{i=1}^{n} \sqrt{1-x_{i}} \end{aligned}$
由柯西不等式
$\begin{aligned} \sum_{i=1}^{n} \sqrt{1-x_{i}}=& \sum_{i=1}^{n}\left(\sqrt{1-x_{i}} \cdot 1\right) \leqslant \sqrt{\sum_{i=1}^{n}\left(1-x_{i}\right)} \cdot \sqrt{\sum_{i=1}^{n} 1}=\sqrt{n-1} \sqrt{n} , &\sum_{i=1}^{n} \sqrt{x_{i}} \leqslant \sqrt{\sum_{i=1}^{n} x_{i} \sqrt{n}}=\sqrt{n} \end{aligned}$
又对于 $n$ 个正数 $y_{1}, \cdots, y_{n}$，由算术-几何-调和平均不等式 $\displaystyle \dfrac{1}{n} \sum\limits_{i=1}^{n} y_{i} \geqslant \sqrt[n]{y_{1} y_{2} \cdots y_{n}} \geqslant \dfrac{n}{\sum\limits_{i=1}^{n} \dfrac{1}{y_{i}}}$
所以 $\displaystyle \sum\limits_{i=1}^{n} \frac{1}{y_{i}} \geqslant \frac{n^{2}}{\sum\limits_{i=1}^{n} y_{i}}$ 把这些不等式合起来，就得到 $\displaystyle \sum\limits_{i=1}^{n} \dfrac{x_{i}}{\sqrt{1-x_{i}}}= \sum_{i=1}^{n} \dfrac{1}{\sqrt{1-x_{i}}}-\sum_{i=1}^{n} \sqrt{1-x_{i}} \geqslant \dfrac{n^{2}}{\sum\limits_{i=1}^{n} \sqrt{1-x_{i}}}-\sum\limits_{i=1}^{n} \sqrt{1-x_{i}} \geqslant \\\dfrac{n^{2}}{\sqrt{n-1} \sqrt{n}}-\sqrt{n-1} \sqrt{n}=\dfrac{n^{2}-(n-1) n}{\sqrt{n-1} \sqrt{n}}=\dfrac{\sqrt{n}}{\sqrt{n-1}} \geqslant\dfrac{1}{\sqrt{n-1}} \sum\limits_{i=1}^{n} \sqrt{x_{i}}$
由柯西不等式
$\begin{aligned} \sum_{i=1}^{n} \sqrt{1-x_{i}}=& \sum_{i=1}^{n}\left(\sqrt{1-x_{i}} \cdot 1\right) \leqslant \sqrt{\sum_{i=1}^{n}\left(1-x_{i}\right)} \cdot \sqrt{\sum_{i=1}^{n} 1}=\sqrt{n-1} \sqrt{n} , &\sum_{i=1}^{n} \sqrt{x_{i}} \leqslant \sqrt{\sum_{i=1}^{n} x_{i} \sqrt{n}}=\sqrt{n} \end{aligned}$
又对于 $n$ 个正数 $y_{1}, \cdots, y_{n}$，由算术-几何-调和平均不等式 $\displaystyle \dfrac{1}{n} \sum\limits_{i=1}^{n} y_{i} \geqslant \sqrt[n]{y_{1} y_{2} \cdots y_{n}} \geqslant \dfrac{n}{\sum\limits_{i=1}^{n} \dfrac{1}{y_{i}}}$
所以 $\displaystyle \sum\limits_{i=1}^{n} \frac{1}{y_{i}} \geqslant \frac{n^{2}}{\sum\limits_{i=1}^{n} y_{i}}$ 把这些不等式合起来，就得到 $\displaystyle \sum\limits_{i=1}^{n} \dfrac{x_{i}}{\sqrt{1-x_{i}}}= \sum_{i=1}^{n} \dfrac{1}{\sqrt{1-x_{i}}}-\sum_{i=1}^{n} \sqrt{1-x_{i}} \geqslant \dfrac{n^{2}}{\sum\limits_{i=1}^{n} \sqrt{1-x_{i}}}-\sum\limits_{i=1}^{n} \sqrt{1-x_{i}} \geqslant \\\dfrac{n^{2}}{\sqrt{n-1} \sqrt{n}}-\sqrt{n-1} \sqrt{n}=\dfrac{n^{2}-(n-1) n}{\sqrt{n-1} \sqrt{n}}=\dfrac{\sqrt{n}}{\sqrt{n-1}} \geqslant\dfrac{1}{\sqrt{n-1}} \sum\limits_{i=1}^{n} \sqrt{x_{i}}$
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