Example 10. Let $P$ be a point inside an acute $\triangle ABC$, and the feet of the perpendiculars from $P$ to the sides $BC$, $CA$, $AB$ be $D$, $E$, $F$ respectively. Find (and prove) the point $P$ that minimizes $PD^2 + PE^2 + PF^2$.

Make the translation and retain the original text's line breaks and format, output the translation result directly.
Let the area of $\triangle ABC$ be $S$, $AB=c$,
$$
\begin{array}{l}
BC=a, CA=b, PD=x, PE=y, PF \\
=z, w=PD^{2}+PE^{2}+PF^{2}=x^{2}+y^{2}+z^{2},
\end{array}
$$

Then we have
$$
\begin{aligned}
S^{2} & =(S \triangle \triangle PB+S \triangle BP C+S \triangle CP A)^{2} \\
& =\frac{1}{4}(a x+b y+c z)^{2} \\
& \leqslant \frac{1}{4}\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right).
\end{aligned}
$$

Thus, $w \geqslant \frac{4 S^{2}}{a^{2}+b^{2}+c^{2}}$.
Equality holds if and only if $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\lambda=\frac{2 S}{a^{2}+b^{2}+b^{2}}$,
$$
w_{\mathrm{min}}=\frac{4 S^{2}}{a^{2}+\bar{b}^{2}+c^{2}}.
$$

Point $P$ can be constructed within the original acute $\triangle ABC$. First, draw a line parallel to $BC$ such that the distance between the two parallel lines is $\lambda a$. Then, draw a line parallel to $AC$ such that the distance between the two parallel lines is $\lambda b$. The intersection point of these two parallel lines within the acute $\triangle ABC$ is the point $P$ we are looking for.