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? 
 A: First note that $\dfrac{A}2, \dfrac{B}2, \dfrac{C}2 \in (0, \pi/2)$. Now consider the function $f(x) = \tan^2(x)$. For $x \in (0,\pi/2)$, $f(x)$ is convex. Hence, we have
$$\dfrac{f(x_1) + f(x_2) + f(x_3)}3 \geq f \left(\dfrac{x_1 + x_2 + x_3}3 \right)$$
Take $x_1 = \dfrac{A}2$, $x_2 = \dfrac{B}2$ and $x_3 = \dfrac{C}2$. We have $x_1 + x_2 + x_3 = \dfrac{A+B+C}2 = \dfrac{\pi}2$.
Hence, we get
$$\dfrac{f(A/2) + f(B/2) + f(C/2)}3 \geq f \left(\dfrac{\pi}6 \right)$$
This gives us
$$\dfrac{\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2)}3 \geq \tan^2(\pi/6) = \dfrac13$$
Hence,
$$\tan^2(A/2) + \tan^2(B/2) + \tan^2(C/2) \geq 1$$

If $f(x)$ is convex, a $n$-gon with vertices on the curve will always lie on the upper side of the curve (In fact, this is the definition of convexity. A quick way to check if a function is convex is to check if its second derivative (if it exists) is non-negative.) and hence the center of mass of the $n$-gon will also lie on the upper side of the curve. In our case, $n=3$.
A: 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)
A: 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, $$\sum (\tan \frac A2-\tan \frac B2)^2\ge0$$
$$2\left(\tan ^2\frac A2+\tan^2\frac B2+\tan^2\frac C2\right)\ge2\sum \tan \frac A2\tan \frac B2$$
