Prove the local inequality first:
$$\tan B \tan C \geqslant \cot ^{2} \frac{A}{2}$$

Since
$$\begin{array}{l} \tan A \tan B=\frac{\cos (A-B)-\cos (A+B)}{\cos (A+B)+\cos (A-B)}= \\ \frac{\cos (A-B)+\cos C}{-\cos C+\cos (A-B)}= \\ {\left[1+\frac{2 \cos C}{\cos (A-B)-\cos C}\right] \geqslant} \\ {\left[1+\frac{2 \cos C}{1-\cos C}\right]=\frac{1+\cos C}{1-\cos C}=\cot ^{2} \frac{C}{2}} \\ \tan B \tan C \geqslant \cot ^{2} \frac{A}{2} \\ \sum \tan B \tan C \geqslant \sum \cot ^{2} \frac{A}{2} \end{array}$$
Thus,

Therefore,
the equality in (9) holds if and only if $\triangle A B C$ is an equilateral triangle.