Example 5 Let $P$ be a point inside $\triangle A B C$, and let $D, E, F$ be the feet of the perpendiculars from $P$ to $B C, C A, A B$ respectively. Determine the point $P$ such that $P D \times P E \times P F$ is maximized.
Let the three interior angles of $\triangle ABC$ be $A, B, C$, and their opposite sides be $a, b, c$. Let the area of $\triangle ABC$ be $S$ (the notation will be the same in the following questions, and will not be repeated).

Let $PD=x, PE=y, PF=z$. Connect $AP, BP, CP$. It is easy to see that $S_{1}(\text{area of } \triangle PBC) = \frac{1}{2} a x, S_{2}(\text{area of } \triangle PCA) = \frac{1}{2} b y, S_{3}(\text{area of } \triangle PAB) = \frac{1}{2} c z$.

Thus, we have $a x + b y + c z = 2(S_{1} + S_{2} + S_{3}) = 2S = \text{constant}$. By the AM-GM inequality, we get
$$\begin{array}{c}
a x \cdot b y \cdot c z \leqslant \left(\frac{a x + b y + c z}{3}\right)^{3} = \left(\frac{2 S}{3}\right)^{3} \\
x y z \leqslant \frac{8 S^{3}}{27 a b c} .
\end{array}$$

The equality holds if and only if $a x = b y = c z$. This means that $S_{1} = S_{2} = S_{3} = \frac{1}{3} S$ maximizes $x y z$. At this time, $P$ is the centroid of $\triangle ABC$.