Question

In $△ABC$, $\tan A+\tan B+\tan C\ge n\sqrt{3}$, where A, B, C are acute angles then find n.

Answer 1

In $△ABC$, $\tan A+\tan B+\tan C=\tan A\tan B\tan C$

Also, $\frac{\tan A+\tan B+\tan C}{3}\ge \sqrt[3]{3\tan A\tan B\tan C}$ [since A.M.$\ge$G.M.]

$\tan A\tan B\tan C\ge 3\sqrt[3]{3\tan A\tan B\tan C}$

${\tan }^{2}A{\tan }^{2}B{\tan }^{2}C\ge 27$

$\tan A\tan B\tan C\ge 3\sqrt{3}$

$\tan A+\tan B+\tan C\ge 3\sqrt{3}$
Ans: 3