16. 2 (IMO 22) Given that $P$ is a point inside $\triangle A B C$, perpendiculars from $P$ to $B C$, $C A$, and $A B$ intersect at points $D$, $E$, and $F$ respectively. Find all points $P$ such that $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}$ is minimized.
16.2 Let $P D=x, P E=y, P F=z$, and the area of $\triangle A B C$ be $S$. The problem is to find the minimum value of $\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}=\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$ under the condition $2 S=a x+b y+c z$. By the Cauchy-Schwarz inequality,
$$
(a x+b y+c z)\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right) \geqslant(a+b+c)^{2} \text {, so } \frac{a}{x}+\frac{b}{y}+\frac{c}{z} \geqslant \frac{(a+b+c)^{2}}{2 S} \text {. }
$$

Equality holds in the above inequality if and only if $x=y=z$, i.e., when $P$ is the incenter of $\triangle A B C$.