15. Let $a_{1}, a_{2}, \cdots, a_{n}$ be given real numbers, not all zero. If the inequality $r_{1}\left(x_{1}-a_{1}\right)+$ $r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}-a_{n}\right) \leqslant \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}-\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}$, holds for all real numbers $x_{1}, x_{2}, \cdots, x_{n}$, find the values of $r_{1}, r_{2}, \cdots, r_{n}$.
15. Let $x_{i}=0(i=1,2, \cdots, n)$, then $r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n} \geqslant$ $\sqrt{a_{1}^{2}+a^{2}+\cdots+a_{n}^{2}}$. Also, let $x_{i}=2 a_{i}(i=1,2, \cdots, n)$, then $r_{1} a_{1}+r_{2} a_{2}+\cdots+$ $r_{n} a_{n} \leqslant \sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}$. Therefore,
$$
r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}=\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} .
$$

Let $x_{i}=r_{i}(i=1,2, \cdots, n)$, then $r_{1}\left(r_{1}-a_{1}\right)+r_{2}\left(r_{2}-a_{2}\right)+\cdots+r_{n}\left(r_{n}-\right.$
$$
\left.a_{n}\right) \leqslant \sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}}-\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots a_{n}^{2}} \text {. }
$$

Substituting (1) into (2), we have $r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2} \leqslant \sqrt{r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2}}$, i.e.,
$$
r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2} \leqslant 1 \text {. }
$$

According to the Cauchy-Schwarz inequality, we have
$$
\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right)^{2} \leqslant\left(r_{1}^{2}+r_{2}^{2}+\cdots r_{n}^{2}\right)\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}\right) .
$$
Substituting (1) into (4), we get $r_{1}^{2}+r_{2}^{2}+\cdots+r_{n}^{2} \geqslant 1$.

From (3) and (5), we have $r_{1}^{2}+r_{2}^{2}+\cdots r_{n}^{2}=1$.
The condition for equality in (4) is $r_{2}=\lambda a_{k}(k=1,2, \cdots n)$, substituting into (6) gives $\lambda=$ $\frac{1}{\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}}$, thus $r_{i}=\frac{a_{i}}{\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}}(i=1,2, \cdots, n)$
Substituting (7) into the given inequality and using (4), we have $r_{1}\left(x_{1}-a_{1}\right)+r_{2}\left(x_{2}-a_{2}\right)+\cdots+r_{n}\left(x_{n}\right.$
$$
\begin{array}{l}
\left.-a_{n}\right)=\left(r_{1} x_{1}+r_{2} x_{2}+\cdots+r_{n} x_{n}\right)-\left(r_{1} a_{1}+r_{2} a_{2}+\cdots+r_{n} a_{n}\right)= \\
\frac{a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}}{\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}}-\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} \leqslant \sqrt{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}- \\
\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}} . \text { Thus, we can determine } r_{i}=\frac{a_{i}}{\sqrt{a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}}}(i=1,2, \cdots, n),
\end{array}
$$
which is the desired result.