Problem - 3874
In $\triangle{ABC}$, show that
$$\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}\le\frac{\sqrt{3}}{9}$$
Applying the AM-GM inequality on the conclusion of %%HREF%%3873%%:
\begin{align*}
1 &= \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}\\
&\ge 3\sqrt[3]{\Big(\tan\frac{A}{2}\tan\frac{B}{2}\Big)\Big(\tan\frac{B}{2}\tan\frac{C}{2}\Big)\Big(\tan\frac{C}{2}\tan\frac{A}{2}\Big)}\\
&=3\sqrt[3]{\Big(\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}\Big)^2}
\end{align*}
$$\therefore\quad\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}\le\frac{\sqrt{3}}{9}$$