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
$$
\begin{array}{l}
\quad r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+ \\
r_{n}\left(x_{n}-a_{n}\right) \\
\leqslant \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}} \\
\quad-\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} \quad \text { (1) }
\end{array}
$$

holds for any real numbers $x_{1}, x_{2}, \cdots, x_{n}$, find the values of $r_{1}$, $r_{2}, \cdots, r_{n}$.
Let $x_{1}=x_{2}=\cdots=x_{n}=0$, from (1) we have
$$
\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \\
\quad \geqslant \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}
\end{array}
$$

Take $x_{1}=2 x_{1}, x_{2}=x_{2}, \cdots, x_{n}=2 a_{n}$, from (1) we also get
$$
\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \\
\leqslant \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} .
\end{array}
$$

From (2) and (3), we obtain
$$
\begin{array}{l}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \\
=\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} .
\end{array}
$$

By the Cauchy-Schwarz inequality,
$$
\begin{array}{c}
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \\
\leqslant \sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}} \\
\cdot \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} .
\end{array}
$$

Since $a_{1}, a_{2}, \cdots, a_{n}$ are not all zero, from (2) and (5) we can conclude.
$$
\sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}} \geqslant 1 \text {. }
$$