4. Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of the following expression:
$$
S=\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} .
$$
(22nd All-Soviet Union Olympiad)
4. 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 $\frac{1}{2}\left(\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}\right) \geqslant y^{2}, \frac{1}{2}\left(\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}}\right) \geqslant z^{2}, \frac{1}{2}\left(\frac{z^{2} x^{2}}{y^{2}}+\frac{x^{2} y^{2}}{z^{2}}\right) \geqslant x^{2}$, adding these three inequalities, we get $\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}} \geqslant x^{2}+y^{2}+z^{2}$. Therefore, $S^{2} \geqslant 3\left(x^{2}+y^{2}+z^{2}\right)=3$, which means $S \geqslant \sqrt{3}$. When $x=y=z=\frac{\sqrt{3}}{3}$, the condition $x^{2}+y^{2}+z^{2}=1$ is satisfied, and $S=\sqrt{3}$, hence the minimum value of $S$ is $\sqrt{3}$.