Example $7 \triangle A B C$ is a known acute triangle, and the regular $\triangle D E F$ is circumscribed around $\triangle A B C$. Find the maximum value of the area of $\triangle D E F$.
(1983, Wuhu City, Anhui Province High School Mathematics Competition)
Solution: As shown in Figure 7, let $\angle A$ be the largest angle of $\triangle ABC$. From $S_{\text{SEEF}} = \frac{\sqrt{3}}{4} EF^2$, we know that $S_{\text{CDEF}}$ is maximized when $EF$ is maximized. Since $EF$ is uniquely determined by $\angle BAF$, we can choose $\angle BAF = \theta$ as the independent variable, then
$$
\begin{aligned}
\angle ABF &= 120^{\circ} - \theta, \\
\angle ACE &= (\theta + \angle A) - 60^{\circ}.
\end{aligned}
$$

By the Law of Sines, we get
$$
\begin{aligned}
EF &= AF + AE \\
&= \frac{c \sin (120^{\circ} - \theta)}{\sin 60^{\circ}} + \frac{b \sin (\theta + \angle A - 60^{\circ})}{\sin 60^{\circ}} \\
&= \frac{2 \sqrt{3} k}{3} \cdot \sin (\theta + \varphi).
\end{aligned}
$$

where, $k = \left\{\left[c \cos 60^{\circ} + b \cos (\angle A - 60^{\circ})\right]^2 \right.$
$$
\begin{aligned}
& + \left[c \sin 60^{\circ} + b \sin (\angle A - 60^{\circ})\right]^2 \zeta^{\frac{1}{2}} \\
& = \sqrt{c^2 + b^2 + 2bc \cos (120^{\circ} - \angle A)}, \\
\varphi &= \arccos \frac{c \cos 60^{\circ} + b \cos (\angle A - 60^{\circ})}{k}
\end{aligned}
$$

is an acute angle.
When $\theta = 90^{\circ} - \varphi$, $EF$ reaches its maximum value $\frac{2 \sqrt{3} k}{3}$, at this time $\theta$ is also an acute angle, so $60^{\circ} < \angle A + \theta < 180^{\circ}$, and the corresponding $\triangle DEF$ is indeed the circumcircle equilateral triangle of $\triangle ABC$. Therefore, the maximum value of $S_{\text{SGEF}}$ is
$$
\frac{\sqrt{3} k^2}{3} = \frac{\sqrt{3}}{3} \left[b^2 + c^2 + 2bc \cos (120^{\circ} - \angle A)\right].
$$