Given the known condition $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$ and the structure of the inequality to be proven, it is not difficult to think of using the Cauchy inequality to first perform a relaxation, obtaining $\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant \sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}}$. Then, by constructing a recursive relationship, we can prove the strengthened inequality of the original inequality. By the Cauchy inequality and $\sum_{i=1}^{n} \frac{1}{a_{i}} \leqslant 1$, we get
$$\left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} \leqslant\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)\left[\sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}}\right] \leqslant \sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}}$$

Therefore, to prove the original inequality, it suffices to prove the following inequality:
$$\sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant \frac{1}{2}.$$
$$\frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}} .$$

To this end, we can first prove the following stronger proposition:
$$\sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant \frac{1}{2} \cdot\left[\frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}-\frac{1}{a_{n+1}\left(a_{n+1}-1\right)+x^{2}}\right],$$
where $a_{n+1}>a_{n}$ is a positive integer.

Let $b_{n}=\frac{1}{2} \cdot\left[\frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}-\frac{1}{a_{n+1}\left(a_{n+1}-1\right)+x^{2}}\right]$. By $a_{k+1} \geqslant a_{k}+1$, we get
$$\begin{aligned} b_{k}-b_{k-1}= & \frac{1}{2} \cdot\left[\frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}-\frac{1}{a_{k+1}\left(a_{k+1}-1\right)+x^{2}}\right] \\ & -\frac{1}{2} \cdot\left[\frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}-\frac{1}{a_{k}\left(a_{k}-1\right)+x^{2}}\right] \\ = & \frac{1}{2} \cdot\left[\frac{1}{a_{k}\left(a_{k}-1\right)+x^{2}}-\frac{1}{a_{k+1}\left(a_{k+1}-1\right)+x^{2}}\right] \\ \geqslant & \frac{1}{2} \cdot\left[\frac{1}{a_{k}\left(a_{k}-1\right)+x^{2}}-\frac{1}{\left(a_{k}+1\right) \cdot a_{k}+x^{2}}\right] \\ = & \frac{a_{k}}{\left(a_{k}^{2}+x^{2}\right)^{2}-a_{k}^{2}} \geqslant \frac{a_{k}}{\left(a_{k}^{2}+x^{2}\right)^{2}}(k \geqslant 2) . \end{aligned}$$
$$\begin{aligned} \text { Also, } b_{1} & =\frac{1}{2} \cdot\left[\frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}-\frac{1}{a_{2}\left(a_{2}-1\right)+x^{2}}\right] \\ & \geqslant \frac{1}{2} \cdot\left[\frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}}-\frac{1}{\left(a_{1}+1\right) a_{1}+x^{2}}\right]=\frac{a_{1}}{\left(a_{1}^{2}+x^{2}\right)^{2}-a_{1}^{2}} \geqslant \frac{a_{1}}{\left(a_{1}^{2}+x^{2}\right)^{2}} \\ \text { Also, } b_{n} & =b_{1}+\sum_{k=2}^{n}\left(b_{k}-b_{k-1}\right) \geqslant \frac{a_{1}}{\left(a_{1}^{2}+x^{2}\right)^{2}}+\sum_{k=2}^{n} \frac{a_{k}}{\left(a_{k}^{2}+x^{2}\right)^{2}}=\sum_{k=1}^{n} \frac{a_{k}}{\left(a_{k}^{2}+x^{2}\right)^{2}} \end{aligned}$$

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