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? 

Hint:
From the identity : $\cos A= 1- 2 \sin^2(\frac{A}{2})$
$\sin^2(\frac{A}{2})=1-\cos A$
$1- \dfrac{b^2+c^2-a^2}{2bc}=\dfrac{2bc-b^2-c^2+a^2}{2bc}$
$= \dfrac{a^2-(b-c)^2}{2bc}=\dfrac{(a+b-c)(a-b+c)}{2bc}$
$a+b-c=\dfrac{(s-b)}{2}$, $a-b+c=\dfrac{(s-c)}{2}$
$2 \sin^2\dfrac{A}{2}=\dfrac{4(s-b)(s-c)}{2bc}$
$\sin^2(\dfrac{A}{2})=\dfrac{(s-b)(s-c)}{bc}$
Similarly you get $\cos^2(\dfrac{A}{2})$.
$\tan^2 \dfrac{A}{2} = \dfrac{(s-a)(s-c)}{(s)(s-a)}$
Add all such functions, you get:
$\dfrac{(s-b)^2(s-c)^2+(s-c)^2(s-a)^2+(s-a)^2(s-b)^2}{(s)(s-a)(s-b)(s-c)}$
To make the calculation simple, use $b+c-a=x$, $c+a-b=y$ and $a+b-c=z$.
$(s-a)=\dfrac{x}{2}$
$(s-b)=\dfrac{y}{2}$
$(s-c)=\dfrac{z}{2}$
$s=\dfrac{x+y+z}{2}$
Now it becomes algebraic expression. You can find the maximum of it by AM-GM inequality!
$\dfrac{(xy)^2+(yz)^2+(zx)^2}{(x+y+z)(xyz)} \ge \dfrac{3(xyz)^{4/3}}{(x+y+z)(xyz)}$ 
$\dfrac{3(xyz)^{1/3}}{(x+y+z)} \ge1$ (AM-GM)
