Problem 1156

When $x^{2} \geqslant a_{1}\left(a_{1}-1\right)$, from $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$, we can get
$$\begin{aligned} \left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} \frac{1}{2 a_{i}|x|}\right)^{2}=\frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} \\ & \leqslant \frac{1}{4 x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}} . \end{aligned}$$

When $x^{2}<a_{1}\left(a_{1}-1\right)$, by the Cauchy inequality, we have
$$\begin{aligned} \left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}}\right) \\ & \leqslant \sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} . \end{aligned}$$

For positive integers $a_{1}<a_{2}<\cdots<a_{n}$, we have $a_{i+1} \geqslant a_{i}+1, i=1,2, \cdots, n-1$, and
$$\begin{array}{l} \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}+\frac{1}{4}\right)^{2}-a_{i}^{2}} \\ =\frac{2 a_{i}}{\left(\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}\right)\left(\left(a_{i}+\frac{1}{2}\right)^{2}+x^{2}\right)} \\ =\frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i}+\frac{1}{2}\right)^{2}+x^{2}} \\ \leqslant \frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}}, \\ \end{array}$$
$$i=1,2, \cdots, n-1 \text {, }$$

Therefore, $\sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant \frac{1}{2} \sum_{i=1}^{n}\left(\frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}}\right)$
$$\leqslant \frac{1}{2} \cdot \frac{1}{\left(a_{1}-\frac{1}{2}\right)^{2}+x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}} .$$