Example 6 (22nd IMO problem) Let $P$ be any point inside triangle $ABC$, and the distances from $P$ to the three sides $BC$, $CA$, $AB$ are $d_{1}$, $d_{2}$, $d_{3}$ respectively. Denote $BC=a$, $CA=b$, $AB=c$. Find the minimum value of $u=\frac{a}{d_{1}} + \frac{b}{d_{2}} + \frac{c}{d_{3}}$. (Example 10 in [1])
Let the area of $\triangle ABC$ be $S$, then $a d_{1}+b d_{2}+c d_{3} = 2 S$. Since $a, b, c, d_{1}, d_{2}, d_{3} \in \mathbf{R}^{+}$, then $u=\frac{a}{d_{1}}+\frac{b}{d_{2}} +\frac{c}{d_{3}}=\frac{a^{2}}{a d_{1}}+\frac{b^{2}}{b d_{2}}+\frac{c^{2}}{c d_{3}} \geqslant \frac{(a+b+c)^{2}}{a d_{1}+b d_{2}+c d_{3}} = \frac{(a+b+c)^{2}}{2 S}$. The equality holds if and only if $\frac{a}{a d_{1}}=\frac{b}{b d_{2}}=\frac{c}{c d_{3}}$, i.e., $d_{1}=d_{2} =d_{3}$, which means $P$ is the incenter of $\triangle ABC$. Therefore, the minimum value of $u=\frac{a}{d_{1}}+ \frac{b}{d_{2}}+\frac{c}{d_{3}}$ is $\frac{(a+b+c)^{2}}{2 S}$.