Example 5 Let $P$ be a point inside $\triangle ABC$, and let $D$, $E$, $F$ be the feet of the perpendiculars from $P$ to $BC$, $CA$, $AB$ respectively: Find all points $P$ that minimize $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$.
(22nd IMO)
Let the area of $\triangle ABC$ be $S$, then
$$
BC \cdot PD + CA \cdot PE + AB \cdot PF = 2S \text{.}
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{\left[\left(\sqrt{\frac{BC}{PD}}\right)^{2}+\left(\sqrt{\frac{CA}{PE}}\right)^{2}+\left(\sqrt{\frac{AB}{PF}}\right)^{2}\right] .} \\
{\left[(\sqrt{BC \cdot PD})^{2}+(\sqrt{CA \cdot PE})^{2}+(\sqrt{AB \cdot PF})^{2}\right]} \\
\geqslant(BC + CA + AB)^{2} . \\
\text{ Therefore, } \frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF} \geqslant \frac{(BC + CA + AB)^{2}}{2S} .
\end{array}
$$

Equality holds if and only if $PD = PE = PF$. In this case, point $P$ is the incenter of $\triangle ABC$.