$\square$ Example 5 Let $a_{1}, a_{2}, \cdots, a_{n}$ be given non-zero real numbers. If the inequality
$$r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}-a_{n}\right) \leqslant \sqrt[m]{x_{1}^{m}+x_{2}^{m}+\cdots+x_{n}^{m}}$$
$-\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$ (where $m \geqslant 2, m \in \mathbf{N}^{*}$) holds for all real numbers $x_{1}, x_{2}, \cdots$,
$x_{n}$, find the values of $r_{1}, r_{2}, \cdots, r_{n}$. (A generalization of Problem 1 from the 3rd CMO)
Substitute $x_{i}=0(i=1,2, \cdots, n)$ into the original inequality, we get
$$-\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right) \leqslant-\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$

That is,
$$r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \geqslant \sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$

Substitute $x_{i}=2 a_{i}(i=1,2, \cdots, n)$ into the original inequality, we get
$$r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \leqslant \sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$

From (1) and (2), we have
$$r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}=\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$

By Hölder's inequality, we get
$$\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}
\end{array}$$
$$\begin{array}{l}
\leqslant\left(r^{\frac{n}{-1}}+\frac{a}{r^{-1}}+\cdots+r_{n}^{\frac{n}{-1}}\right)^{\frac{n-1}{m}}
\end{array}$$

Therefore,
$$\left(r_{\mathrm{T}}^{\frac{m}{-1}}+r_{\frac{n}{2}}^{\frac{n}{1}}+\cdots+r_{n}^{\frac{m}{n-1}}\right)^{\frac{--1}{m}} \geqslant 1$$

That is,
$$r_{\mathrm{T}}^{\frac{n}{-1}}+r_{2}^{\frac{n}{2}}+\cdots+r_{n}^{\frac{n}{-1}} \geqslant 1$$

Substitute (3) into the original inequality, we get
$$r_{1} x_{1}+r_{2} x_{2}+\cdots+r_{n} x_{n} \leqslant \sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}$$

From (5) and (6), we have
$$r_{1}^{\frac{n}{-1}}+r_{2}^{\frac{m}{-1}}+\cdots+r_{n}^{\frac{n}{-1}}=1$$

From (7), we get the inequality (4) holds if and only if
$$\begin{array}{l}
r_{1}^{\frac{1}{n^{-1}}}: a_{1}=r^{\frac{1}{n-1}}: a_{2}=\cdots=r^{\frac{1}{n^{-1}}}: a_{n} \\
=\sqrt[m]{r_{1}^{\frac{m}{1}}+r_{2}^{\frac{m}{-1}}+\cdots+r_{n}^{\frac{m}{-1}}}: \sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}} \\
=1: \sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}
\end{array}$$

That is, when $r_{i}=\left(\frac{a_{i}}{\sqrt[m]{a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}}}\right)^{m-1}(i=1,2, \cdots, n)$, the equality holds.

Substitute the obtained $r_{i}(i=1,2, \cdots, n)$ into the left side of the original inequality, and use the generalized Cauchy inequality, we get
$$\begin{aligned}
& r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}-a_{n}\right) \\
= & \left(r_{1} x_{1}+r_{2} x_{2}+\cdots+r_{n} x_{n}\right)-\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right) \\
\leqslant & \left(\sum_{i=1}^{n} r_{i}^{m-1}\right)^{\frac{m}{m}}\left(\sum_{i=1}^{m} x_{i}^{m}\right)^{\frac{1}{m}}-\sum_{i=1}^{m}\left(\frac{a_{i}}{\sqrt{\sum_{i=1}^{n} a_{i}^{m}}}\right)^{m-1} \cdot a_{i} \\
= & \left(\sum_{i=1}^{m} x_{i}^{m}\right)^{\frac{1}{m}}-\left(\sum_{i=1}^{m} a_{i}^{m}\right)^{\frac{1}{m}}
\end{aligned}$$

This means that the obtained $r_{i}(i=1,2, \cdots, n)$ can make the original inequality hold for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$.