9- 53 Let $x, y, z$ be positive numbers and $x^{2}+y^{2}+z^{2}=1$, find
$$
\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}
$$

the minimum value.
[Solution] Let $x=x_{1}, y=x_{2}, z=x_{3}$. By the Cauchy-Schwarz inequality,
$$
\begin{aligned}
1=\left(\sum_{i=1}^{3} x_{i}^{2}\right)^{2} & =\left(\sum_{i=1}^{3} \sqrt{\frac{x_{i}}{1-x_{i}^{2}}} \cdot x_{i}^{\frac{3}{2}} \sqrt{1-x_{i}^{2}}\right)^{2} \\
& \leqslant\left(\sum_{i=1}^{3} \frac{x_{i}}{1-x_{i}^{2}}\right)\left(\sum_{i=1}^{3} x_{i}^{3}\left(1-x_{i}^{2}\right)\right),
\end{aligned}
$$

By the AM-GM inequality, we have
$$
\frac{2 x_{i}^{2}}{3 \sqrt{3}}+x_{i}^{5} \geqslant 3 \sqrt[3]{\frac{x_{i}^{2}}{3 \sqrt{3}} \cdot \frac{x_{i}^{2}}{3 \sqrt{3}} \cdot x_{i}^{5}}=x_{i}^{3},
$$

Thus,
$$
\frac{2}{3 \sqrt{3}}+\sum_{i=1}^{3} x_{i}^{5} \geqslant \sum_{i=1}^{3} x_{i}^{3},
$$

which implies $\sum_{i=1}^{3} x_{i}^{3}\left(1-x_{i}^{2}\right) \leqslant \frac{2}{3 \sqrt{3}}$.
From (1), we can deduce that
$$
\sum_{i=1}^{3} \frac{x_{i}}{1-x_{i}^{2}} \geqslant \frac{3 \sqrt{3}}{2}.
$$

On the other hand, when $x=y=z=\frac{\sqrt{3}}{3}$,
$$
\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}=\frac{3 \sqrt{3}}{2}.
$$

Therefore, the minimum value sought is $\frac{3 \sqrt{3}}{2}$.