In a acute angled triangle, we have $\tan(A)\cdot\tan(B)\cdot\tan(C) \geq 3\sqrt{3}$ How to show:

In a acute angled $\triangle  \ ABC$ show that $$\tan(A) \cdot \tan(B)\cdot \tan(C) \geq 3\sqrt{3}$$

Any ideas?

$$A+B=\pi-C$$

\begin{align*}
&\tan  (A+B)= \tan  (\pi-C)\\
&(\tan  A+ \tan  B)/(1-\tan  A \tan B)= (\tan  \pi- \tan  C)/(1+\tan  \pi \tan  C)=-\tan C\\
&(\tan  A+ \tan  B)= -\tan  C(1-\tan  A \tan  B)\\
&\tan  A + \tan  B= -\tan  C+ \tan  A \tan  B \tan  C\\
&\tan  A + \tan  B+ \tan  C= \tan  A \tan  B \tan  C
\end{align*}

