Example 3. In an equilateral $\triangle ABC$, a point $P$ inside the triangle is such that the segments $PA=6, PB=8, PC=10$. The integer closest to the area of $\triangle ABC$ is ().
(A) 159
(B) 131
(C) 95
(D) 79
(E) 50
(1967, 18th American High School Mathematics Examination)
From the previous example, we know that $a=6, b=8, c=10$, and the key to the problem is to find $u, v, z$.
And $S_{\triangle A B C}=2\left(S_{\triangle 1}\right.$
$$
\begin{array}{l}
\left.+S_{\triangle 1}+S_{\triangle 1}\right)+\left(S_{\triangle T}\right. \\
\left.+S_{\triangle 2}+S_{\triangle G}\right)= \\
2 S_{\triangle H F G}+\frac{\sqrt{3}}{4}\left(u^{2}+v^{2}\right. \\
\left.+w^{2}\right) .
\end{array}
$$

Since $\triangle H F G$ is a right triangle $\left(\because 10^{2}=6^{2}+\right.$ $\left.8^{2}\right)$,
$$
\text { then } \begin{aligned}
S_{\triangle H F G} & =\frac{1}{2} \cdot 8 \cdot 6 \\
& =24 .
\end{aligned}
$$

In $\triangle O H I, \triangle O I G$,
$\triangle O G H$, using the cosine theorem respectively, we have
(Note $\angle H O I=\angle I O G$
$$
\begin{array}{l}
\left.=\angle G O H=120^{\circ}\right) ; \\
\left\{\begin{array}{l}
u^{2}+w^{2}+u w=10^{2}, \\
v^{2}+w^{2}+v w=6^{2}, \\
u^{2}+v^{2}+u v=8^{2} .
\end{array}\right.
\end{array}
$$
(1) + (2) + (3) gives
$$
2\left(u^{2}+v^{2}+w^{2}\right)+u v+v z w+z u u=200,
$$

By the rearrangement inequality
$$
u v+v w+w u \leqslant u^{2}+v^{2}+w^{2} \text {, }
$$

we have $2\left(u^{2}+v^{2}+w^{2}\right) \leqslant 200$
$$
\leqslant 3\left(u^{2}+v^{2}+w^{2}\right),
$$

which means $\frac{200}{3} \leqslant u^{2}+v^{2}+w^{2} \leqslant 100$, then
$$
48+\frac{\sqrt{3}}{4} \cdot \frac{290}{3} \leqslant S_{\triangle A B C} \leqslant 48+\frac{\sqrt{3}}{4} .
$$
100 ,
which means $76.9 \leqslant S_{\triangle A B^{*}} \leqslant 91.3$. Therefore, the answer is D.