Trigonometry Intermediate

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*} $$\therefore\quad\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}\le\frac{\sqrt{3}}{9}$$

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