Example 10 Let real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy
$$
\sum_{i=1}^{n} x_{i}^{2}=1 \quad(n \geqslant 3) \text {. }
$$

Find $\min _{1 \leqslant i<j \leqslant n}\left\{\left|x_{i}-x_{j}\right|\right\}$'s maximum value.
Explanation: By symmetry, without loss of generality, assume $x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{n}$, and let $\min _{1 \leqslant i<j \leqslant n}\left\{\left|x_{i}-x_{j}\right|\right\}=m$.
By the minimality of $m$, we have
$$
x_{j}-x_{i} \geqslant(j-i) m(1 \leqslant i<j \leqslant m) \text {. }
$$

Then $\sum_{1 \leqslant i<j \leqslant n}\left(x_{i}-x_{j}\right)^{2}$
$$
\geqslant m^{2} \sum_{1 \leqslant i<j \leqslant n}(i-j)^{2}=\frac{m^{2} n^{2}\left(n^{2}-1\right)}{12} \text {. }
$$

On the other hand, by a well-known identity, we have
$$
\begin{aligned}
& \sum_{1 \leqslant i<j \leqslant n}\left(x_{i}-x_{j}\right)^{2} \\
= & (n-1) \sum_{i=1}^{n} x_{i}^{2}-2 \sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j} \\
= & n \sum_{i=1}^{n} x_{i}^{2}-\left(\sum_{i=1}^{n} x_{i}\right)^{2} \leqslant n .
\end{aligned}
$$

Therefore, $\frac{m^{2} n^{2}\left(n^{2}-1\right)}{12} \leqslant n$, which means
$$
m \leqslant \sqrt{\frac{12}{n\left(n^{2}-1\right)}} .
$$

It is easy to see that when $\sum_{i=1}^{n} x_{i}=0$, and $\left\{x_{i}\right\}$ forms an arithmetic sequence, the equality holds.
Thus, the maximum value of $m$ is $\sqrt{\frac{12}{n\left(n^{2}-1\right)}}$.
$$
\begin{array}{l}
\text { Note: } \sum_{1 \leqslant i<j \leqslant n}(i-j)^{2} \\
=\sum_{1 \leqslant i<j \leqslant n}\left(i^{2}+j^{2}\right)-2 \sum_{1 \leqslant i<j \leqslant n} i j \\
=(n-1) \sum_{i=1}^{n} i^{2}-2 \sum_{1 \leqslant i<j \leqslant n} i j \\
=n \sum_{i=1}^{n} i^{2}-\left(\sum_{i=1}^{n} i\right)^{2}=\frac{n^{2}\left(n^{2}-1\right)}{12} .
\end{array}
$$