Example 3. In an equilateral $\triangle ABC$, a point $P$ inside the triangle is connected to the three vertices with $PA=6, PB=8, PC=10$. The integer closest to the area of $\triangle ABC$ is ( ).
(A) 159
(B) 131
(C) 95
(D) 79
(E) 50
From the previous example, we know that $a=6, b=8, c=10$, the key to the problem lies in finding $u, z, zw$.
$$48+\frac{\sqrt{3}}{4} \cdot \frac{290}{3} \leqslant S_{\triangle A B C} \leqslant 48+\frac{\sqrt{3}}{4} \cdot 100,$$
i.e., $76.9 \leqslant S_{\triangle A B E} \leqslant 91.3$. Therefore, the answer is D.
If it is not too far-fetched, the following problem can also be considered a variation or extension of the initial problem.
$$\begin{array}{l}
\quad \text { and } S_{\triangle A B C}=2\left(S_{\triangle I}\right. \\
\left.+S_{\triangle I}+S_{\triangle I}\right)+\left(S_{\triangle(j)}\right. \\
\left.+\quad S_{\triangle W}+S_{\triangle(3)}\right)= \\
2 S_{\triangle H F G}+\frac{\sqrt{3}}{4}\left(u^{2}+v^{2}\right. \\
\left.+w^{2}\right) .
\end{array}$$

Since $\triangle H F G$ is a right triangle $\left(\because 10^{2}=6^{2}+\right.$ $\left.8^{2}\right)$,

then $S_{\triangle H F C}=\frac{1}{2} \cdot 8 \cdot 6$
$$=24$$

In $\triangle O H I, \triangle O I G$,
$\triangle O G H$, using the cosine rule respectively (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
$$2\left(u^{2}+v^{2}+w^{2}\right)+u v+v w+w v u=200,$$

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

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

Thus, $\frac{200}{3} \leqslant u^{2}+v^{2}+w^{2} \leqslant 100$, then