Given positive numbers $x, y, z$ satisfying $x^{2}+y^{2}+z^{2}=1$.
(1) Find the minimum value of $\frac{x^{2} y}{z^{2}}+\frac{y^{2} z}{x^{2}}+\frac{z^{2} x}{y^{2}}$;
(2) Find the minimum value of $\frac{x^{2} y^{3}}{z^{4}}+\frac{y^{2} z^{3}}{x^{4}}+\frac{z^{2} x^{3}}{y^{4}}$.
Solution: (1) Let $s=\frac{x^{2} y}{z^{2}}+\frac{y^{2} z}{x^{2}}+\frac{z^{2} x}{y^{2}}$, then
$$
\begin{array}{l}
s^{2}=\frac{x^{4} y^{2}}{z^{4}}+\frac{y^{4} z^{2}}{x^{4}}+\frac{z^{4} x^{2}}{y^{4}}+2\left(\frac{x^{3}}{y}+\frac{y^{3}}{z}+\frac{z^{3}}{x}\right) \\
\geqslant 3\left(\frac{x^{3}}{y}+\frac{y^{3}}{z}+\frac{z^{3}}{x}\right) . \\
\text { Also, } \frac{x^{3}}{y}+x y \geqslant 2 x^{2}, \frac{y^{3}}{z}+z y \geqslant 2 y^{2}, \frac{z^{3}}{x}+x z \geqslant 2 z^{2}, \text { and }
\end{array}
$$
$x y+y z+z x \leqslant x^{2}+y^{2}+z^{2}$, hence
$$
\begin{array}{c}
\frac{x^{3}}{y}+\frac{y^{3}}{z}+\frac{z^{3}}{x} \\
\geqslant 2\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x) \\
\geqslant x^{2}+y^{2}+z^{2}=1 .
\end{array}
$$

Equality holds if and only if $x=y=z=\frac{\sqrt{3}}{3}$. Therefore, $s^{2} \geqslant 3, s \geqslant \sqrt{3}$.
Thus, when $x=y=z=\frac{\sqrt{3}}{3}$, $\frac{x^{2} y}{z^{2}}+\frac{y^{2} z}{x^{2}}+\frac{z^{2} x}{y^{2}}$ has a minimum value of $\sqrt{3}$.
$$
\begin{array}{l}
\text { (2) Let } s=\frac{x^{2} y^{3}}{z^{4}}+\frac{y^{2} z^{3}}{x^{4}}+\frac{z^{2} x^{3}}{y^{4}}, \text { similarly to (1) we get } \\
s^{2} \geqslant 3\left(\frac{x^{5}}{z^{2} y}+\frac{y^{5}}{x^{2} z}+\frac{z^{5}}{y^{2} x}\right) . \\
\text { Also, } \frac{x^{5}}{z^{2} y}+z^{2}+x y \geqslant 3 x^{2}, \frac{y^{5}}{x^{2} z}+x^{2}+y z \geqslant 3 y^{2}, \\
\frac{z^{5}}{y^{2} x}+y^{2}+z x \geqslant 3 z^{2},
\end{array}
$$

and $x y+y z+z x \leqslant x^{2}+y^{2}+z^{2}$, hence
$$
\begin{array}{l}
\frac{x^{5}}{z^{2} y}+\frac{y^{5}}{x^{2} z}+\frac{z^{5}}{y^{2} x} \\
\geqslant 3\left(x^{2}+y^{2}+z^{2}\right)-\left(x^{2}+y^{2}+z^{2}\right)-(x y+y z+z x) \\
\geqslant x^{2}+y^{2}+z^{2}=1 .
\end{array}
$$

Equality holds if and only if $x=y=z=\frac{\sqrt{3}}{3}$. Therefore, $s^{2} \geqslant 3, s \geqslant \sqrt{3}$.
Thus, when $x=y=z=\frac{\sqrt{3}}{3}$, $\frac{x^{2} y^{3}}{z^{4}}+\frac{y^{2} z^{3}}{x^{4}}+\frac{z^{2} x^{3}}{y^{4}}$ has a minimum value of $\sqrt{3}$.
(Zhang Jun, Zhaoyang Middle School, Xinghua City, Jiangsu Province, 225700)