Example 14 Let the three sides of $\triangle ABC$ be $a, b, c$ with corresponding altitudes $h_{a}$, $h_{b}$, $h_{c}$, and the radius of the incircle of $\triangle ABC$ be $r=2$. If $h_{a}+h_{b}+h_{c}=18$, find the area of $\triangle ABC$.
Given that $h_{a}+h_{b}+h_{c}=18$, we have $(a+b+c)\left(h_{a}+h_{b}+\right.$ $\left.h_{c}\right)=18(a+b+c)=9(a+b+c) r$.

By the Cauchy-Schwarz inequality, we get
$(a+b+c)\left(h_{a}+h_{b}+h_{c}\right) \geqslant\left(\sqrt{a h_{a}}+\sqrt{b h_{b}}+\sqrt{c h_{c}}\right)^{2}$.
Also, the area of $\triangle A B C$ is $S=\frac{1}{2} a h_{a}=\frac{1}{2} b h_{b}=\frac{1}{2} c h_{c}=\frac{1}{2}(a+b+c) r$,
so $9 \times 2 S \geqslant(\sqrt{2 S}+\sqrt{2 S}+\sqrt{2 S})^{2}=18 S$,
with equality if and only if $\frac{h_{a}}{a}=\frac{h_{b}}{b}=\frac{h_{c}}{c}$.
Since $a h_{a}=b h_{b}=c h_{c}$,
we have $a=b=c$, and $\triangle A B C$ is an equilateral triangle.
Given $r=\frac{\sqrt{3}}{6} a=2$, we find $a=4 \sqrt{3}$.
Thus, the area of $\triangle A B C$ is $S=\frac{\sqrt{3}}{4} a^{2}=\frac{\sqrt{3}}{4} \times(4 \sqrt{3})^{2}=12 \sqrt{3}$.