Example 2 As shown in Figure 1, given that $P$ is a point inside $\triangle ABC$, and $D$, $E$, $F$ are the feet of the perpendiculars from $P$ to $BC$, $CA$, and $AB$ respectively. Find the point $P$ that minimizes $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$.
(22nd IMO)
Let $B C+C A+A B=l$, and the area of $\triangle A B C$ be $S$. Then
$$
B C \cdot P D+C A \cdot P E+A B \cdot P F=2 S .
$$

Construct the function $f(x)=\frac{1}{x}$.
Since $\frac{B C}{l}+\frac{C A}{l}+\frac{A B}{l}=1$, by Jensen's inequality, we have
$$
\begin{array}{l}
\frac{B C}{l} \cdot \frac{1}{P D}+\frac{C A}{l} \cdot \frac{1}{P E}+\frac{A B}{l} \cdot \frac{1}{P F} \\
\geqslant \frac{1}{\frac{B C}{l} \cdot P D+\frac{C A}{l} \cdot P E+\frac{A B}{l} \cdot P F},
\end{array}
$$

which means
$$
\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F} \geqslant \frac{l^{2}}{2 S} \text {. }
$$

The equality holds if and only if $P D=P E=P F$, so the point $P$ that minimizes $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$ is the incenter of $\triangle A B C$.