Example 19 (Problem 22 of the 1988 All-Union Mathematical Olympiad) 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:
$$
S=\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} .
$$
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-Geometric Mean Inequality, we have
$$
\begin{array}{l}
\frac{1}{2}\left(\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}\right) \geqslant y^{2}, \\
\frac{1}{2}\left(\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}}\right) \geqslant z^{2} \\
\frac{1}{2}\left(\frac{z^{2} x^{2}}{y^{2}}+\frac{x^{2} y^{2}}{z^{2}}\right) \geqslant x^{2},
\end{array}
$$
Adding (1), (2), and (3) yields
$$
\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}} \geqslant x^{2}+y^{2}+z^{2},
$$

Thus,
$$
S^{2} \geqslant 3\left(x^{2}+y^{2}+z^{2}\right)=3, \quad S \geqslant \sqrt{3}.
$$

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