Example 6.13 Given that $m$ is a positive even number, and real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\sum_{i=1}^{n} x_{i}^{m}=1$, find $\max \min _{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|$.
Proof (Sun Shibao) First, we prove a lemma:
Lemma If $m$ is a positive even number, then for any real numbers $a, b$, we have
$$\frac{a^{m}+b^{m}}{2} \geqslant\left(\frac{a-b}{2}\right)^{m}$$

Proof of the lemma: Let $x, y \geqslant 0$, by the power mean inequality, we have $\frac{x^{m}+y^{m}}{2} \geqslant\left(\frac{x+y}{2}\right)^{m}$, thus $\frac{|a|^{m}+|b|^{m}}{2} \geqslant\left(\frac{|a|+|b|}{2}\right)^{m}$. Also, $|a|+|b| \geqslant|a-b|$, so we have
$$\frac{a^{m}+b^{m}}{2} \geqslant\left(\frac{|a-b|}{2}\right)^{m} \geqslant\left(\frac{a-b}{2}\right)^{m}$$

The lemma is proved. Now, let's return to the original inequality. Without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$, and let $S=\sum_{i=1}^{n}(n+1-2 i)^{m}$, $t=\min _{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|$. If $i<j$, then
$$\begin{aligned}
\left|x_{i}-x_{j}\right|= & \left|\left(x_{i}-x_{i+1}\right)+\left(x_{i+1}-x_{i+2}\right)+\cdots+\left(x_{j-1}-x_{j}\right)\right|= \\
& \left|x_{i}-x_{i+1}\right|+\left|x_{i+1}-x_{i+2}\right|+\cdots+ \\
& \left|x_{j-1}-x_{j}\right| \geqslant(j-i) t
\end{aligned}$$

That is, $\left|x_{i}-x_{n+1-i}\right| \geqslant|n+1-2 i| t, i=1,2, \cdots, n$. By the lemma, we get
$$\frac{x_{i}^{m}+x_{n+1-i}^{m}}{2} \geqslant\left(\frac{x_{i}-x_{n+1-i}}{2}\right)^{m} \geqslant\left[\frac{(n+1-2 i) t}{2}\right]^{m}=\frac{(n+1-2 i)^{m} t^{m}}{2^{m}}$$

Given $\sum_{i=1}^{n} x_{i}^{m}=1$, we know
$$1=\sum \frac{x_{i}^{m}+x_{n+1-i}^{m}}{2} \geqslant \frac{t^{m}}{2^{m}} \sum_{i=1}^{n}(n+1-2 i)^{m}$$

Thus, $t \leqslant \frac{2}{\sqrt[m]{S}}$, equality holds if and only if the sequence $x_{n}$ is an arithmetic sequence, and $x_{i}+x_{n+1-i}=0$, that is, the sum of the first $n$ terms of the arithmetic sequence $x_{n}$, $s_{n}=x_{1}+x_{2}+\cdots+x_{n}=0$. Therefore,
$$\max \min _{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|=\frac{2}{\sqrt{\sum_{i=1}^{n}(n+1-2 i)^{m}}}$$