Example 3 As shown in Figure 2.10.3, there is a fixed point $P$ inside $\angle A$, and a line passing through $P$ intersects the two sides at $B, C$. When does $\frac{1}{P B}+\frac{1}{P C}$ reach its maximum value?
Draw $A D \perp B C$ at $D$. Let $A D=h$, then
$$
\begin{aligned}
\frac{1}{P B}+\frac{1}{P C}= & \frac{h}{2}\left(\frac{1}{S_{\triangle A B P}}+\frac{1}{S_{\triangle A C P}}\right)= \\
& \frac{h}{2} \cdot \frac{2 A B \cdot A C \cdot \sin \angle B A C}{A B \cdot A C \cdot A P^{2} \cdot \sin \angle P A B \cdot \sin \angle P A C}= \\
& h \cdot \frac{\sin \angle B A C}{A P^{2} \cdot \sin \angle P A B \cdot \sin \angle P A C} \leqslant \\
& \frac{\sin \angle B A C}{A P \cdot \sin \angle P A B \cdot \sin \angle P A C}
\end{aligned}
$$

Since $P$ is a fixed point inside $\angle A$, $A B, \angle P A B, \angle P A C$ are all constants, so the last term $\frac{\sin \angle B A C}{A P \cdot \sin \angle P A B \cdot \sin \angle P A C}$ is a constant. When $A P \perp B C$, $A D$ coincides with $A P$, and the inequality holds as an equality. Therefore, when $A P \perp B C$, $\frac{1}{P B}+\frac{1}{P C}$ achieves its maximum value.