6・97 Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, try to find the minimum value of the following expression
$$S=\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} .$$
［Solution］ Squaring the expression, we get
$$S^{2}=\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}}+2\left(x^{2}+y^{2}+z^{2}\right)$$

By the arithmetic mean and geometric mean inequality, we have
$$\begin{array}{l}
\frac{1}{2}\left(\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}\right) \geqslant y^{2} \\
\frac{1}{2}\left(\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}}\right) \geqslant z^{2} \\
\frac{1}{2}\left(\frac{z^{2} x^{2}}{y^{2}}+\frac{x^{2} y^{2}}{z^{2}}\right) \geqslant x^{2}
\end{array}$$
(1) + (2) + (3) gives
$$\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}} \geqslant x^{2}+y^{2}+z^{2}$$

Thus,
$$\begin{array}{l}
S^{2} \geqslant 3\left(x^{2}+y^{2}+z^{2}\right)=3 \\
S \geqslant \sqrt{3}
\end{array}$$

Furthermore, when $x=y=z=\frac{\sqrt{3}}{3}$, the given condition $x^{2}+y^{2}+z^{2}=1$ is satisfied, and $S=\sqrt{3}$. Therefore, the minimum value of $S$ is $\sqrt{3}$.