9. (16 points) Let real numbers $x, y, z > 0$. Find the minimum value of the following two expressions:
(1) $\max \left\{x, \frac{1}{y}\right\} + \max \left\{y, \frac{2}{x}\right\}$;
(2) $\max \left\{x, \frac{1}{y}\right\} + \max \left\{y, \frac{2}{z}\right\} + \max \left\{z, \frac{3}{x}\right\}$.
(1) Original expression $\geqslant x+\frac{2}{x} \geqslant 2 \sqrt{2}$, when $x=\sqrt{2}$ and $\frac{1}{y} \leqslant x$ with $y \leqslant \frac{2}{x}$ hold simultaneously, i.e., $\frac{\sqrt{2}}{2} \leqslant y \leqslant \sqrt{2}$, the original expression reaches its minimum value $2 \sqrt{2}$.
(2) Note that, original expression $\geqslant x+\frac{2}{z}+\max \left\{z, \frac{3}{x}\right\}$.
Take an undetermined coefficient $\lambda \in(0,1)$, then
$$
\begin{array}{l}
\text { Original expression } \geqslant x+\frac{2}{z}+\lambda z+(1-\lambda) \frac{3}{x} \\
\geqslant 2 \sqrt{3(1-\lambda)}+2 \sqrt{2 \lambda} .
\end{array}
$$

For the equality to hold, we should have
$$
\left\{\begin{array}{l}
\frac{1}{x} \leqslant y \leqslant \frac{2}{z}, \\
z=\frac{3}{x}=\sqrt{\frac{2}{\lambda}}, \\
x=\sqrt{3(1-\lambda)} .
\end{array}\right.
$$

Solving gives $\lambda=\frac{2}{5}$.
Substituting in, the minimum value of the original expression is $2 \sqrt{5}$, which is achieved when $x=\frac{3 \sqrt{5}}{5}$, $\frac{\sqrt{5}}{3} \leqslant y \leqslant \frac{2 \sqrt{5}}{5}, z=\sqrt{5}$.