15. (GBR) ${ }^{\mathrm{IMO}}$ Find the point $P$ inside the triangle $A B C$ for which
$$
\frac{B C}{P D}+\frac{C A}{P E}+\frac{A B}{P F}
$$
is minimal, where $P D, P E, P F$ are the perpendiculars from $P$ to $B C, C A$, $A B$ respectively.
15. Set $B C=a, C A=b, A B=c$, and denote the area of $\triangle A B C$ by $P$, and $a / P D+b / P E+c / P F$ by $S$. Since $a \cdot P D+b \cdot P E+c \cdot P F=2 P$, by the Cauchy-Schwarz inequality we have
$$
2 P S=(a \cdot P D+b \cdot P E+c \cdot P F)\left(\frac{a}{P D}+\frac{b}{P E}+\frac{c}{P F}\right) \geq(a+b+c)^{2} \text {, }
$$
with equality if and only if $P D=P E=P F$, i.e., $P$ is the incenter of $\triangle A B C$. In that case, $S$ attains its minimum:
$$
S_{\text {min }}=\frac{(a+b+c)^{2}}{2 P} .
$$