Example 5 Let $P$ be a point inside an acute $\triangle ABC$, and the feet of the perpendiculars from $P$ to the sides $BC$, $CA$, and $AB$ are $D$, $E$, and $F$ respectively. Find (and prove) the point $P$ that minimizes $\overline{P D}^{2}+\overline{P E}^{2}+\overline{P F}^{2}$.
$(1990$, Zhejiang Province High School Mathematics Summer Camp)
Solution: As shown in Figure 5, let $S$ be the area of $\triangle ABC$, with side lengths $a, b, c$, and $PD$
$$
\begin{aligned}
=x, PE & =y, PF=z, \\
w & =\overline{PD^{2}}+\overline{PE^{2}}+\overline{PF^{2}} \\
& =x^{2}+y^{2}+z^{2} .
\end{aligned}
$$

We have
$$
\begin{aligned}
S^{2}= & \left(S_{\triangle PBC}+S_{\triangle PCA}\right. \\
& \left.+S_{\triangle PAB}\right)^{2} \\
= & \frac{1}{4}(a x+b y+c z)^{2} \\
\leqslant & \frac{1}{4}\left(a^{2}+b^{2}+c^{2}\right)\left(x^{2}+y^{2}+z^{2}\right) .
\end{aligned}
$$

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

Of course, point $P$ can also be constructed as follows: Draw a line parallel to $BC$ intersecting the other two sides, and let its distance from $BC$ be $\lambda a$. Similarly, draw a line parallel to $AC$ and let its distance from $AC$ be $\lambda b$. The intersection point of these two lines within $\triangle ABC$ is point $P$.