Example 4 Let $P$ be a point inside $\triangle ABC$, and draw perpendiculars from $P$ to the three sides $BC$, $CA$, and $AB$, with the feet of the perpendiculars being $D$, $E$, and $F$ respectively. For what position of $P$ is the value of $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$ minimized? (22nd IMO problem)
Solution: Let $B C=a, C A=b$, $A B=c, P D=d_{1}, P E=d_{2}$, $P F=d_{3},$ and the area of $\triangle A B C$ be $\Delta$. Clearly, we have $2 \Delta=a d_{1}+b d_{2}+c d_{3}$.
$$\begin{array}{c}
\therefore \frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}=\frac{a}{d_{1}}+ \\
\frac{b}{d_{2}}+\frac{c}{d_{3}} \geqslant \frac{(a+b+c)^{2}}{\left(a d_{1}+b d_{2}+c d_{3}\right)}=\frac{(a+b+c)^{2}}{2 \Delta} \text {. When }
\end{array}$$

and only when $d_{1}=d_{2}=d_{3}$, the equality holds, i.e., when $P$ is the incenter of $\triangle A B C$, $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$ has the minimum value
$$\frac{(a+b+c)^{2}}{2 \Delta}$$