<s>[INST] Trigonometric Functions And Identities Question . If $A$ , $B$ and $C$ are the angles of a triangle , I have to show that : 
$ \tan^2 \cfrac{A}{2} + \tan^2 \cfrac{B}{2} + \tan^2 \cfrac{C}{2} \ge 1 $ . 
I had only arrived at $A+B+C = \pi $ , thus $\cfrac{A}{2} + \cfrac{B}{2} + \cfrac{C}{2} = \cfrac{\pi}{2} $ 
What to do next? 
 [/INST] As $A+B=\pi-C\implies \frac{A+B}2=\frac\pi2-\frac C2$
So, $$\tan\left(\frac{A+B}2\right)=\tan\left(\frac\pi2-\frac C2\right)$$
$$\frac{\tan \frac A2+\tan \frac B2}{1-\tan \frac A2\tan \frac B2}=\cot\frac C2=\frac1{\tan\frac C2} $$
$$\sum \tan \frac A2\tan \frac B2=1$$
Now, $$\sum (\tan \frac A2-\tan \frac B2)^2\ge0$$
$$2\left(\tan ^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)\ge2\sum \tan \frac A2\tan \frac B2$$
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