9. (14 points) In an acute $\triangle A B C$, it is known that $\angle A=$ $75^{\circ}, A C=b, A B=c$. Find the maximum value of the area of the equilateral triangle circumscribed around $\triangle A B C$.

In an acute $\triangle A B C$, it is known that $\angle A=$ $75^{\circ}, A C=b, A B=c$. Find the maximum value of the area of the equilateral triangle circumscribed around $\triangle A B C$.
9. Let $\triangle D E F$ be the circumscribed equilateral triangle of $\triangle A B C$, as shown in Figure 1.
Let $\angle B A F=\alpha$.
Then, by the Law of Sines, we have
$$
\begin{aligned}
A E & =\frac{b \sin \left(\alpha+15^{\circ}\right)}{\sin 60^{\circ}}, \\
A F & =\frac{c \sin \left(120^{\circ}-\alpha\right)}{\sin 60^{\circ}}
\end{aligned}
$$

Thus, $E F=A E+A F$
$$
\begin{array}{l}
=\frac{2}{\sqrt{3}}\left[\left(b \sin 15^{\circ} \cdot \cos \alpha+b \cos 15^{\circ} \cdot \sin \alpha\right)+\right. \\
\left.\left(\frac{\sqrt{3}}{2} c \cos \alpha+\frac{1}{2} c \sin \alpha\right)\right] \\
=\frac{2}{\sqrt{3}}\left[\left(\frac{c}{2}+b \cos 15^{\circ}\right) \sin \alpha+\right. \\
\left.\left(\frac{\sqrt{3}}{2} c+b \sin 15^{\circ}\right) \cos \alpha\right] \\
\leqslant \frac{2}{\sqrt{3}} \sqrt{\left(\frac{c}{2}+b \cos 15^{\circ}\right)^{2}+\left(\frac{\sqrt{3}}{2} c+b \sin 15^{\circ}\right)^{2}} \\
=\frac{2}{\sqrt{3}} \sqrt{b^{2}+c^{2}+b c\left(\cos 15^{\circ}+\sqrt{3} \sin 15^{\circ}\right)} \\
=\frac{2}{\sqrt{3}} \sqrt{b^{2}+c^{2}+2 b c \cos 45^{\circ}} \\
=\frac{2}{\sqrt{3}} \sqrt{b^{2}+c^{2}+\sqrt{2} b c} \text {. } \\
\end{array}
$$

When $\alpha=\frac{\pi}{2}-\arctan \frac{\frac{\sqrt{3}}{2} c+b \sin 15^{\circ}}{\frac{c}{2}+b \cos 15^{\circ}}$, the equality holds.
Therefore, the maximum area of the circumscribed equilateral triangle of $\triangle A B C$ is
$$
\begin{array}{l}
\frac{\sqrt{3}}{4} E F^{2}=\frac{\sqrt{3}}{4}\left(\frac{2}{\sqrt{3}} \sqrt{b^{2}+c^{2}+\sqrt{2} b c}\right)^{2} \\
=\frac{\sqrt{3}}{3}\left(b^{2}+c^{2}+\sqrt{2} b c\right) .
\end{array}
$$