1. In $\triangle A B C$, $a^{2}+b^{2}+c^{2}=2 \sqrt{3} a b \sin C$, then $\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}=$ $\qquad$
$$
\begin{array}{l}
a^{2}+b^{2}+c^{2}=a^{2}+b^{2}+\left(a^{2}+b^{2}-2 a b \cos C\right)=2 \sqrt{3} a b \sin C \\
\Rightarrow \sqrt{3} a b \sin C+a b \cos C=2 a b \sin \left(C+30^{\circ}\right)=a^{2}+b^{2} \geqslant 2 a b,
\end{array}
$$

The equality holds if and only if $\triangle A B C$ is an equilateral triangle. Therefore, $\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}=\left(\frac{\sqrt{3}}{2}\right)^{3}=\frac{3 \sqrt{3}}{8}$.