If $\tan A + \tan B + \tan C = 3 \sqrt{3}$, then $\triangle ABC$ is equilateral. 
Given $\triangle ABC$ such that 
  $$\tan A + \tan B + \tan C = 3 \sqrt{3}$$
  prove that $\triangle ABC$ is equilateral.

The full process is needed.
 A: First we prove $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
We know that

$A + B + C = 180$

from the angle addition formula, so

$\tan((A + B) + C) = \frac{\tan(A + B) + \tan C}{1−\tan(A + B) \tan C}$
$= \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan A \tan C - \tan B \tan C}$

We know that tan $180 = 0$, and since $A+B+C = 0$, we have

$\tan A + \tan B + \tan C - \tan A \tan B \tan C = 0$
$\tan A + \tan B + \tan C = \tan A \tan B \tan C$

Now we apply the AM-GM inequality, $\frac{a+b+c}3 \ge abc^{\frac13}$. ($x^{\frac13}$ is the cube root)

$\frac{\tan A + \tan B + \tan C}3 \ge (\tan A \tan B \tan C)^{\frac13}$
$\frac{\tan A \tan B \tan C}3 \ge (\tan A \tan B \tan C)^{\frac13}$

We substitute $x$ for $\tan A \tan B \tan C$.

$\frac x3 \ge x^{\frac13}$
$x \ge 3 x^{\frac13}$
$x^3 \ge 27 x$
$x^2 \ge 27$
$x \ge 3 \sqrt 3 $
$\tan A \tan B \tan C >= 3 \sqrt(3)$
$\tan A + \tan B + \tan C >= 3 \sqrt(3)$

The AM-GM inequality only becomes an equality when $x_1 = x_2 = x_3$, as can be seen from this page on wikipedia.
Thus, we know that $\tan A + \tan B + \tan C = 3 \sqrt 3$ only when $\tan A = \tan B = \tan C$. This will only be true when all angle are the same, so for $\tan A + \tan B + \tan C = 3 \sqrt 3$ to be true, the triangle has to be equilateral.
