Example 3 Let $P$ be a moving point inside $\triangle ABC$, and the distances from $P$ to the three sides $a, b, c$ are denoted as $r_{1}, r_{2}, r_{3}$, respectively. Try to find the point $P$ that minimizes the value of $\frac{a}{r_{1}}+\frac{b}{r_{2}}+\frac{c}{r_{3}}$.
(22nd IMO)
Let the area of $\triangle ABC$ be $S$, then $a r_{1}+b r_{2}+c r_{3}=2 S$ is a constant, and the perimeter $a+b+c$ is a constant. Let
$\alpha=\left(\sqrt{a r_{1}}, \sqrt{b r_{2}}, \sqrt{c r_{3}}\right)$,
$\boldsymbol{\beta}=\left(\sqrt{\frac{a}{r_{1}}}, \sqrt{\frac{b}{r_{2}}}, \sqrt{\frac{c}{r_{3}}}\right)$.
Since $(\boldsymbol{\alpha} \cdot \boldsymbol{\beta})^{2} \leqslant|\boldsymbol{\alpha}|^{2}|\boldsymbol{\beta}|^{2}$, i.e.,
$(a+b+c)^{2} \leqslant\left(\frac{a}{r_{1}}+\frac{b}{r_{2}}+\frac{c}{r_{3}}\right)\left(a r_{1}+b r_{2}+c r_{3}\right)$.
Therefore, $\frac{a}{r_{1}}+\frac{b}{r_{2}}+\frac{c}{r_{3}} \geqslant \frac{(a+b+c)^{2}}{2 S}$ is a constant, with equality holding if and only if $r_{1}=r_{2}=r_{3}$.

Thus, the point $P$ that minimizes $\frac{a}{r_{1}}+\frac{b}{r_{2}}+\frac{c}{r_{3}}$ is the incenter of $\triangle ABC$.