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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 :) .

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    When the one angle exceeding $\frac{2\pi}{3}$ is removed, the inequality becomes $\tan {\frac{A}{2}} + \tan {\frac{B}{2}} + \tan{\frac{C}{2}} \geq \sqrt {3} $ – 2017-01-26
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    @S.C.B. Ok Thanks, what about the rest of the question ???? Any ideas ????? – 2017-01-26
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    Do you understand my answer? – 2017-01-26

2 Answers 2

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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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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!