16. 3 (IMO 26 Prelim) For any point $P$ inside $\triangle ABC$, let $D$, $E$, $F$ be the intersections of $AP$, $BP$, $CP$ with $BC$, $CA$, $AB$ respectively. Find the point $P$ such that the area of $\triangle DEF$ is maximized.
16. 3 Let $[A B C]$ denote the area of $\triangle A B C$, $B D=x, C E=y, A F=z$.

We have
$$
\begin{aligned}
{[A B C] } & =\frac{1}{2} a c \sin B, \quad[F B D]=[A B C] \frac{x(c-z)}{a c} \\
{[D E F] } & =[A B C]-[F B D]-[D C E]-[E A F] \\
& =[A B C][1-u(1-w)-v(1-u)-w(1-v)]
\end{aligned}
$$

Here, $u=\frac{x}{a}, v=\frac{y}{b}, w=\frac{z}{c}$.
Thus, the area ratio $F=1-u(1-w)-v(1-u)-w(1-v)=(1-u)(1-v)(1-w)+u v w$
By Ceva's Theorem, $x y z=(a-x)(b-y)(c-z)$, i.e., $u v w=(1-u)(1-v)(1-w)=G$, then $F=2 G$, and when $G$ reaches its maximum value, $F$ also reaches its maximum value. $G^{2}=u(1-u) v(1-v) w(1-w)$
It is easy to deduce that $G \leqslant \frac{1}{8}$, when $u=v=w=\frac{1}{2}$, i.e., when $P$ is the centroid, $[D E F]$ reaches its maximum value $=\frac{1}{4}[A B C]$.