17. Let $x, y, z$ be positive numbers, and $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $S=\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}$.
(1988 Soviet Union Mathematical Olympiad Problem)
17. Since $S$ is a positive number, $S$ and $S^{2}$ attain their minimum values at the same point. Given that $x^{2}+y^{2}+z^{2}=1$, by the AM-GM inequality, we have
$$\begin{aligned}
S^{2}= & \left(\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}\right)^{2}=\frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}}+2\left(x^{2}+y^{2}+z^{2}\right)= \\
& \frac{x^{2} y^{2}}{z^{2}}+\frac{y^{2} z^{2}}{x^{2}}+\frac{z^{2} x^{2}}{y^{2}}+2=\frac{x^{2}}{2}\left(\frac{y^{2}}{z^{2}}+\frac{z^{2}}{y^{2}}\right)+\frac{y^{2}}{2}\left(\frac{x^{2}}{z^{2}}+\frac{z^{2}}{x^{2}}\right)+ \\
& \frac{z^{2}}{2}\left(\frac{y^{2}}{x^{2}}+\frac{x^{2}}{y^{2}}\right)+2 \geqslant x^{2}+y^{2}+z^{2}+2=3
\end{aligned}$$

Therefore, $S$ attains its minimum value $\sqrt{3}$ if and only if $x=y=z=\frac{\sqrt{3}}{3}$.