Let $a_{1}, a_{2}, \cdots, a_{n}$ be given non-negative real numbers, not all zero, $m$ be a given rational number greater than 1, and $r_{1}, r_{2}, \cdots, r_{n}$ be non-negative real numbers such that the inequality
$$
\sum_{k=1}^{n} r_{k}\left(x_{k}-a_{k}\right) \leqslant\left(\sum_{k=1}^{n} x_{k}^{m}\right)^{\frac{1}{m}}-\left(\sum_{k=1}^{n} a_{k}^{m}\right)^{\frac{1}{m}}
$$

holds for any non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$. Find the values of $r_{1}$, $r_{2}, \cdots, r_{n}$.

Obviously, when $m=2$, it is the first problem of the third Winter Camp.
Solution: Let $x_{1}=t \neq a_{1}, x_{2}=a_{2}, x_{3}=a_{3}, \cdots \cdots$ $x_{n}=a_{n}$.
From equation (1) we get
$$
\begin{array}{l}
r_{1}\left(t-a_{1}\right) \\
\leqslant\left(t^{m}+\sum_{k=2}^{n} a_{k}^{m}\right)^{\frac{1}{m}}-\left(a_{1}^{m}+\sum_{k=2}^{n} a_{k}^{m}\right)^{\frac{1}{m}} . \\
\text { Let } b=\sum_{k=2}^{n} a_{k}^{m}, f(t)=\left(t^{m}+b\right)^{\frac{1}{m}} .
\end{array}
$$

Then inequality (2) becomes
$$
r_{1}\left(t-a_{1}\right) \leqslant f(t)-f\left(a_{1}\right),
$$

which holds for all $t \neq a_{1}$.
When $t>a_{1}$, inequality (3) is
$$
r_{1} \leqslant \frac{f(t)-f\left(a_{1}\right)}{t-a_{1}} \text {. }
$$

Let $t \rightarrow a_{1}^{+}$. Then
$$
r_{1} \leqslant f_{+}^{\prime}\left(a_{1}\right)=f^{\prime}\left(a_{1}\right) \text {. }
$$

Similarly, when $t<a_{1}$, inequality (3) is
$$
r_{1} \geqslant f^{\prime}-\left(a_{1}\right)=f^{\prime}\left(a_{1}\right) .
$$

From inequalities (4) and (5) we get
$$
r_{1}=f^{\prime}\left(a_{1}\right)=a_{1}^{m-1}\left(\sum_{k=1}^{n} a_{k}^{m}\right)^{\frac{1-m}{m}} \text {. }
$$

Similarly,
$$
r_{i}=a_{i}^{m-1}\left(\sum_{k=1}^{n} a_{k}^{m}\right)^{\frac{1-m}{m}}(i=1,2, \cdots, n) .
$$

Substituting these values into inequality (1), we only need to prove
$$
\begin{array}{l}
\sum_{k=1}^{n} a_{k}^{m-1}\left(\sum_{l=1}^{n} a_{l}^{m}\right)^{\frac{1-m}{m}}\left(x_{k}-a_{k}\right) \\
\leqslant\left(\sum_{k=1}^{n} x_{k}^{m}\right)^{\frac{1}{m}}-\left(\sum_{k=1}^{n} a_{k}^{m}\right)^{\frac{1}{m}} \\
\Leftrightarrow \sum_{k=1}^{n} a_{k}^{m-1} x_{k} \leqslant\left(\sum_{k=1}^{n} x_{k}^{m}\right)^{\frac{1}{m}}\left(\sum_{k=1}^{n} a_{k}^{m}\right)^{\frac{m-1}{m}} .
\end{array}
$$

The above inequality is the Hölder's inequality.
Note: (1) For the original problem (i.e., $m=2$), the Cauchy inequality is needed to verify inequality (1), which is a special form of Hölder's inequality.
(2) If the right side of inequality (1) is changed to
$$
h\left(\sum_{k=1}^{n} g\left(x_{k}\right)\right)-h\left(\sum_{k=1}^{n} g\left(a_{k}\right)\right) \text {, }
$$

using the same method, we can let
$$
f_{i}(t)=h\left(g(t)+\sum_{k \neq i} g\left(a_{k}\right)\right) .
$$

Then $r_{i}=f_{i}^{\prime}\left(a_{i}\right)=h^{\prime}\left(\sum_{k=1}^{n} g\left(a_{k}\right)\right) g^{\prime}\left(a_{i}\right)$.
Substituting $r_{i}$ back into inequality (1) gives
$$
\begin{array}{l}
\sum_{k=1}^{n} h^{\prime}\left(\sum_{l=1}^{n} g\left(a_{l}\right)\right) g^{\prime}\left(a_{k}\right)\left(x_{k}-a_{k}\right) \\
\leqslant h\left(\sum_{k=1}^{n} g\left(x_{k}\right)\right)-h\left(\sum_{k=1}^{n} g\left(a_{k}\right)\right) .
\end{array}
$$