8-(22nd $I M O$ Problem) Let $P$ be any point inside triangle $A B C$, and let the distances from $P$ to the three sides $B C, C A, A B$ be $d_{1}, d_{2}, d_{3}$ respectively, with $B C=a, C A=b, A B=c$. Find the minimum value of $u=\frac{a}{d_{1}}+\frac{b}{d_{2}}+\frac{c}{d_{3}}$.
Given the area of triangle $A B C$ is $S$, then $a d_{1}+b d_{2}+$ $c d_{3}=2 S$. From this and the Cauchy-Schwarz inequality, we have $2 S \cdot u=\left(a d_{1}+b d_{2}\right.$ $\left.+c d_{3}\right)\left(\frac{a}{d_{1}}+\frac{b}{d_{2}}+\frac{c}{d_{3}}\right) \geqslant(a+b+c)^{2}$, which means $u \geqslant$ $\frac{(a+b+c)^{2}}{2 S}$ (equality holds if and only if $\frac{\sqrt{a d_{1}}}{\sqrt{\frac{a}{d_{1}}}}=\frac{\sqrt{b d_{2}}}{\sqrt{\frac{b}{d_{2}}}}=\frac{\sqrt{c d_{3}}}{\sqrt{\frac{c}{d_{3}}}}$, i.e., $d_{1}=d_{2}=d_{3}$). Therefore, when point $P$ is the incenter of triangle $A B C$, $u$ attains its minimum value $\frac{(a+b+c)^{2}}{2 S}$.