给定正整数 $n \geqslant 2$，设正整数 $a_{i}(i=1,2, \cdots, n)$ 满足 $a_1<a_{2}<\cdots<a_{n}$ 以及 $\displaystyle \sum\limits_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$．求证：对任意实数 $x$，有 $\displaystyle \left(\sum\limits_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}$
【难度】
【出处】
2004第19届CMO试题
【标注】
【答案】
略
【解析】
当 $x^{2} \geqslant a_{1}\left(a_{1}-1\right)$ 时，由 $\displaystyle \sum\limits_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$ 可得
$\displaystyle \left(\sum\limits_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant\left(\sum_{i=1}^{n} \frac{1}{2 a_{i}} \frac{1}{|x|}\right)^{2}=\frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} \leqslant\frac{1}{4 x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}$
当 $x^{2}<a_{1}\left(a_{1}-1\right)$ 时，由柯西不等式得 $\displaystyle \begin{aligned}\left(\sum\limits_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant &\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}}\right) \leqslant \sum_{i=1}^{n} \frac{a^{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \end{aligned}$
对于正整数 $a_{1}<a_{2}<\cdots<a_{n}$，有 $a_{i+1} \geqslant a_{i}+1, i=1,2, \cdots, n-1$ 且
$\begin{aligned} \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant & \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}+\frac{1}{4}\right)^{2}-a_{i}^{2}}= \frac{2 a_{i}}{\left(\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}\right)\left(\left(a_{i}+\frac{1}{2}\right)^{2}+x^{2}\right)}= \frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}} (i=1,2, \cdots, n-1) \end{aligned}$
所以 $\displaystyle \sum\limits_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant\frac{1}{2} \sum_{i=1}^{n}\left(\frac{-1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}}\right) \leqslant\frac{1}{2} \cdot \frac{1}{\left(a_{1}-\frac{1}{2}\right)^{2}+x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}$
$\displaystyle \left(\sum\limits_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant\left(\sum_{i=1}^{n} \frac{1}{2 a_{i}} \frac{1}{|x|}\right)^{2}=\frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} \leqslant\frac{1}{4 x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}$
当 $x^{2}<a_{1}\left(a_{1}-1\right)$ 时，由柯西不等式得 $\displaystyle \begin{aligned}\left(\sum\limits_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant &\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}}\right) \leqslant \sum_{i=1}^{n} \frac{a^{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \end{aligned}$
对于正整数 $a_{1}<a_{2}<\cdots<a_{n}$，有 $a_{i+1} \geqslant a_{i}+1, i=1,2, \cdots, n-1$ 且
$\begin{aligned} \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant & \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}+\frac{1}{4}\right)^{2}-a_{i}^{2}}= \frac{2 a_{i}}{\left(\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}\right)\left(\left(a_{i}+\frac{1}{2}\right)^{2}+x^{2}\right)}= \frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}} (i=1,2, \cdots, n-1) \end{aligned}$
所以 $\displaystyle \sum\limits_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant\frac{1}{2} \sum_{i=1}^{n}\left(\frac{-1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}}\right) \leqslant\frac{1}{2} \cdot \frac{1}{\left(a_{1}-\frac{1}{2}\right)^{2}+x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}$
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