TITLE: In a acute angled triangle, we have $\tan(A)\cdot\tan(B)\cdot\tan(C) \geq 3\sqrt{3}$
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: These may be useful:

Since $\triangle ABC$ is acute, we have $\tan(A),\tan(B),\tan(C)$ positive.
By A.M-G.M you have $$\displaystyle \frac{\tan(A)+\tan(B)+\tan(C)}{3} \geq \sqrt[3]{\tan(A)\cdot\tan(B)\cdot\tan(C)}$$
$\text{The equality holds if the triangle is}$ $\textbf{equilateral.}$