Proof (1) As shown in the figure,
Let the areas of $\triangle P A B, \triangle P B C, \triangle P A C$ be $S_{1}, S_{2}, S_{3}$, and the areas of their projections on the base $\triangle O A B, \triangle O B C, \triangle O C A$ be $S^{\prime}{ }_{1}, S^{\prime}{ }_{2}, S^{\prime}{ }_{3}$, respectively. The area of $\triangle A B C$ is $S$, and $P A = a, P B = b, P C = c$. It is easy to prove that $O$ is the orthocenter of $\triangle A B C$, and the line $C D \perp A B$ through $O$, thus $P D \perp A B, P C \perp P D$.
$$\begin{array}{l} \because \frac{1}{3} S_{1} c=\frac{1}{3} S_{2} b=\frac{1}{3} S_{3} a=\frac{1}{3} S h, \\ \therefore S_{1}=\frac{h}{c} S, S_{2}=\frac{h}{b} S, S_{3}=\frac{h}{c} S . \end{array}$$

Since $P C \perp P D$, we have
$$S_{1}=\frac{S^{\prime} 1}{\cos \angle P D C}=\frac{S^{\prime} 1}{\sin \angle P C O}=\frac{S^{\prime} 1}{\frac{h}{c}} .$$

Similarly, $S_{2}=\frac{S^{\prime} z}{\frac{h}{b}}, S_{3}=\frac{S^{\prime} 3}{\frac{h}{a}}$. Substituting these into (1) we get
$S^{\prime}{ }_{2}=\frac{h^{2}}{c^{2}} S, S^{\prime}{ }_{2}=\frac{h^{2}}{b^{2}} S, S^{\prime}{ }_{3}=\frac{h^{2}}{a^{2}} S$.
Adding these three equations, and noting that $S^{\prime}{ }_{1}+S^{\prime}{ }_{2}+S^{\prime}{ }_{3}=S$, we obtain
$$\frac{1}{h^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} \text {. }$$
(2) We have
$$V_{M-P A C}+V_{M-P A B}+V_{M-P B C}=V_{C-P A B} .$$

That is, $\frac{1}{6} a c y+\frac{1}{6} a b z+\frac{1}{6} b c x=\frac{1}{6} a b c$.
$$\therefore \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \text {. }$$

Using the arithmetic mean inequality, we get
$$\frac{x y z}{a b c} \leqslant\left(\frac{\frac{x}{a}+\frac{y}{b}+\frac{z}{c}}{3}\right)^{3}=\left(\frac{1}{3}\right)^{3}=\frac{1}{27},$$

with equality holding if and only if $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
$$\therefore\left(\frac{x y z}{a b c}\right)_{\max }=\frac{1}{27} \text {. }$$

From the conclusion of (1), we can derive some interesting formulas:
(i) $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=1$;
(ii) $\cos ^{2} \alpha_{1}+\cos ^{2} \beta_{1}+\cos ^{2} \gamma_{1}=1$;
(iii) $S^{2}=S_{1}^{2}+S_{2}^{2}+S_{3}^{2}$.

Here, $\alpha, \beta, \gamma$ are the angles between the lateral edges and the base, and $\alpha_{1}, \beta_{1}, \gamma_{1}$ are the angles between the lateral faces and the base.