Example $6 P$ is a point inside $\triangle A B C$, and $D, E, F$ are the feet of the perpendiculars from $P$ to the sides $B C, C A, A B$ respectively. Find all points $P$ that minimize $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$.
Solution: It is easy to know that
$$
S_{\triangle B C}=\frac{1}{2} B C \cdot P D+\frac{1}{2} C A \cdot P E+\frac{1}{2} A B \cdot P F \text {. }
$$

According to the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F} \\
=\frac{B C \cdot P D+C A \cdot P E+A B \cdot P F}{2 S_{\triangle A B C}}\left(\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}\right) \\
\geqslant \frac{1}{2 S_{\triangle A B C}}\left(\sqrt{B C \cdot P D} \cdot \sqrt{\frac{B C}{P D}}+\right. \\
\left.\sqrt{C A \cdot P E} \cdot \sqrt{\frac{C A}{P E}}+\sqrt{A B \cdot P F} \cdot \sqrt{\frac{A B}{P F}}\right)^{2} \\
=\frac{(B C+C A+A B)^{2}}{2 S_{\triangle A B C}} \text { (a constant). }
\end{array}
$$

The equality in (2) holds if and only if
$$
\begin{array}{l}
\frac{\sqrt{\frac{B C}{P D}}}{\sqrt{B C \cdot P D}}=\frac{\sqrt{\frac{C A}{P E}}}{\sqrt{C A \cdot P E}}=\frac{\sqrt{\frac{A B}{P F}}}{\sqrt{A B \cdot P F}} \\
\Leftrightarrow P D=P E=P F
\end{array}
$$

Therefore, when $P$ is the incenter of $\triangle A B C$, $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$ achieves its minimum value.