Question

Prove that a triangle ABC is equilateral if and only if $\tan A+\tan B+\tan C=3\sqrt{3}$.

Answer 1

$\Delta$ equilateral $\Rightarrow A=B=C={60}^o$
$\therefore \tan A+\tan B+\tan C=3\sqrt{3}$
Converse $S_1=3\sqrt{3}=S_3$ in a triangle
Since $\tan A\tan B\tan C$ is $+$ive, each much be $+$ive as two of them cannot be $-$ive since two angles of $\Delta$ are not obtuse
$\therefore A.M.\ge G.M.$
$\Rightarrow \frac{\sum \tan A}{3}\ge (\tan A\tan B\tan C{)}^{1/3}$
$\therefore S_1^3\ge 27S_3$ or $S_1^3\ge 27S_1$ or $S_1^2\ge 27$
$\therefore S_1\ge 3\sqrt{3}$ but $S_1=3\sqrt{3}$ given.
Hence equality occurs when
$\tan A=\tan B=\tan C$ or $A=B=C$
$\therefore \Delta$ is equilateral.