Prove: The inequality can be rewritten as follows
$$\sum_{i=1}^{n}\left(\frac{1}{n-1+a_{i}^{2}}-\frac{1}{n}\right) \leq 0$$

or equivalently,
$$\sum_{i=1}^{n} \frac{a_{i}^{2}-1}{n-1+a_{i}^{2}} \geq 0$$

Assuming $a_{1} \geq a_{2} \geq \cdots \geq a_{n}$, according to the problem, we have
$$\sum_{i=1}^{n} \frac{a_{i}^{2}-1}{a_{i}}=0$$

Moreover, note that
$$\frac{a_{i}}{n-1+a_{i}^{2}}-\frac{a_{j}}{n-1+a_{j}^{2}}=\frac{\left(n-1-a_{i} a_{j}\right)\left(a_{i}-a_{j}\right)}{\left(n-1+a_{i}^{2}\right)\left(n-1+a_{j}^{2}\right)}$$

Thus, when $a_{i} a_{j} \leq n-1$ for $i \neq j$, we have
$$\sum_{i=1}^{n} \frac{a_{i}^{2}-1}{n-1+a_{i}^{2}} \geq \frac{1}{n}\left(\sum_{i=1}^{n} \frac{a_{i}^{2}-1}{a_{i}}\right)\left(\sum_{i=1}^{n} \frac{a_{i}}{n-1+a_{i}^{2}}\right)=0$$

It remains to prove the case when $a_{1} a_{2} \geq n-1$. For $n \geq 3$, by the Cauchy-Schwarz inequality, we have
$$\frac{a_{1}^{2}}{n-1+a_{1}^{2}}+\frac{a_{2}^{2}}{n-1+a_{2}^{2}} \geq \frac{\left(a_{1}+a_{2}\right)^{2}}{2(n-1)+a_{1}^{2}+a_{2}^{2}} \geq 1 \Rightarrow \sum_{i=1}^{n} \frac{a_{i}^{2}}{n-1+a_{i}^{2}} \geq 1 \Rightarrow \sum_{i=1}^{n} \frac{1}{n-1+a_{i}^{2}} \leq 1$$

For $n=1$ and $n=2$, the inequality becomes an equality. For $n \geq 3$, equality holds if and only if $a_{1}=\cdots=a_{n}=1$.