G44 (22-1, UK) 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 the expression $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$.


It is known that
\[
\begin{aligned}
& B C \cdot P D + C A \cdot P E + A B \cdot P F \\
= & 2\left(S_{\triangle A B P} + S_{\triangle B C P} + S_{\triangle C A P}\right) \\
= & 2 S_{\triangle A B C} .
\end{aligned}
\]

Let \(\frac{B C}{P D} + \frac{C A}{P E} + \frac{A B}{P F} = x\). Then, by the Cauchy-Schwarz inequality, we have
\[
\begin{aligned}
2 x S_{\triangle A B C} & = \left(\frac{B C}{P D} + \frac{C A}{P E} + \frac{A B}{P F}\right)(B C \cdot P D + C A \cdot P E + A B \cdot P F) \\
& \geqslant (B C + C A + A B)^{2} \\
& = C_{\triangle A B C}^{2} .
\end{aligned}
\]
where \(C_{\triangle A B C}\) is the perimeter of \(\triangle A B C\).
The equality in the Cauchy-Schwarz inequality holds if and only if \(P D = P E = P F\). Therefore, the expression achieves its minimum value \(\frac{C_{\triangle A B C}^{2}}{2 S_{\triangle A B C}}\) when \(P\) is the incenter of \(\triangle A B C\).