<s>[INST] $\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 :) .
 [/INST] Let $ \tan\dfrac A2 =x, \tan\dfrac B2=y, \tan\dfrac C2=z,\tan  \dfrac{A}{2}  + \tan \dfrac{B}{2} + \tan\dfrac{C}{2} = w$. WLOG $x \ge \sqrt{3}$.
Note that we have the identity $\tan\left(\dfrac{A+B}2\right)=\tan\left(\dfrac{\pi- C}2\right)$
$$\implies\dfrac{x+y}{1-xy}=\dfrac1{z}$$
$$\implies xy+yz+zx=1 \le \frac{(y+z)^2}{4}+x(w-x)=\frac{(w-x)^2}{4}+x(w-x) $$
This gives us $$w \ge \sqrt{4+4x^2}-x \ge 4- \sqrt{3}$$
From the fact that $\sqrt{4+4x^2}-x $ is an increasing function. 
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