1 Let $a_{1}, a_{2}, \cdots, a_{n}$ be given real numbers, not all zero, and $r_{1}, r_{2}, \cdots, r_{n}$ be real numbers. If the inequality
$$
\begin{array}{l}
r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}-a_{n}\right) \leqslant \\
\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}-\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}
\end{array}
$$

holds for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, find the values of $r_{1}, r_{2}, \cdots, r_{n}$.
Given the assumption, for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, the inequality holds:
$$
\begin{array}{l}
r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}-a_{n}\right) \leqslant \\
\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}-\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}
\end{array}
$$

Substituting $x_{k}=2 a_{k}$ and $x_{k}=0 (k=1,2, \cdots, n)$ into equation (1), we get:
$$
\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \leqslant \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} \\
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \geqslant \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}
\end{array}
$$

Therefore, $r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}=\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}$. Thus, inequality (1) becomes:
$$
r_{1} x_{1}+r_{2} x_{2}+\cdots+r_{n} x_{n} \leqslant \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}
$$

From equation (4) and the Cauchy-Schwarz inequality:
$$
\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+r_{3} a_{3}+\cdots+r_{n} a_{n} \leqslant \\
\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} \cdot \sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}}
\end{array}
$$

It follows that $1 \leqslant \sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}}$. Since equation (4) holds for any real numbers $x_{1}, \cdots, x_{n}$, let $x_{k}=r_{k} (k=1,2, \cdots, n)$, then $\sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}} \leqslant 1$, thus $r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}=1$.
From the above conclusions, we have:
$$
\begin{aligned}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}= & \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}= \\
& \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} . \\
& \sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}}
\end{aligned}
$$

Since the Cauchy-Schwarz inequality becomes an equality, there exists a real number $t$ such that $r_{k}=t a_{k} (k=1,2, \cdots, n)$. Therefore:
$$
t a_{1}^{2}+t a_{2}^{2}+\cdots+t a_{n}^{2}=\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}
$$

Solving for $t$, we get:
$$
t=\frac{1}{\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}}
$$

Thus, $r_{k}=\frac{a_{k}}{\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}}, k=1,2, \cdots, n$.
Note: The key to this problem is that the inequality (1) holds for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, which implies that inequality (4) holds for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, and the equality $r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}=\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}$. As required by the problem, it is not necessary to verify that such $r_{1}, r_{2}, \cdots, r_{n}$ ensure that inequality (1) holds.