$\tan {\frac{A}{2}} + \tan {\frac{B}{2}} +\tan{\frac{C}{2}} \geq 4 - \sqrt {3} $ 
In a triangle $ABC$ with one angle exceeding $\frac {2}{3} \pi$, prove that
$\tan  {\frac{A}{2}}  + \tan {\frac{B}{2}} + \tan{\frac{C}{2}}  \geq  4 - \sqrt {3} $

I tried expanding that half angle, applying AM-GM on various sets, using Sine rule and Napier's Analogy, but without success.
Can anyone provide a hint ?
Also, how does the left hand side of the inequality behave when the condition of one angle exceeding $\frac {2} {3 }\pi$  is removed?
Thanks in advance :) .

Let $\gamma\geq\frac{2\pi}{3}$ and $\tan\frac{\gamma}{4}=x$.
Hence, $\frac{1}{\sqrt3}\leq x<\frac{1}{\sqrt2}$ and since $\tan$ is a convex function on $\left[0,\frac{\pi}{2}\right)$, by Jensen we obtain:
$$\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}\geq2\tan\frac{\alpha+\beta}{4}=2\tan\left(\frac{\pi}{4}-\frac{\gamma}{4}\right)=\frac{2(1-x)}{1+x}.$$
Thus, it remains to prove that 
$$\frac{2(1-x)}{1+x}+\frac{2x}{1-x^2}\geq4-\sqrt3$$ or
$$(\sqrt3x-1)\left(x+\frac{1}{4+3\sqrt3}\right)\geq0,$$ 
which is obvious.
Done!
