If A, B, C are the angles of a triangle then show that 
i) $$\sin A + \sin B + \sin C \le \frac{{3\sqrt 3 }}{2}$$
II) $${\tan ^2}\frac{A}{2} + {\tan ^2}\frac{B}{2} + {\tan ^2}\frac{C}{2} > 1.$$

Solution

$$1.$$We'll choose to set the function f(x) = sin x, on the bracket $$(0,\Pi)$$(we don't have to forget that we are working in a triangle, where the sum of the angles is $$180^\circ$$, meaning $$\Pi$$, if we're measuring the angles in radians.)
$$f^{'}(x) = cos x, f^{''}(x) = - sin x < 0$$, so the function is concave and we we'll apply the Jensen's inequality, which says that:
$$f[\cfrac{A+B+C}{3}] > \cfrac{f(A) + f(B) + f(C)}{3}$$
Working in a triangle, $$A + B + C = 180^\circ$$ and $$\cfrac{(A +  B  + C)}{3} = \cfrac{180}{3} = 60^\circ$$
$$f(60) = sin 60^\circ = \cfrac{\sqrt3}{2}$$
$$\cfrac{f(A) + f(B) + f(C)}{3} = \cfrac{sin A + sin B + sin C}{3}$$
Putting the results again in Jensen's inequality:
$$\cfrac{sin A + sin B + sin C}{3} < \cfrac{\sqrt3}{2}$$
$$sin A + sin B + sin C < \cfrac{3(\sqrt3)}{2} \space q.e.d.$$

$$2.$$ $$tan(\cfrac{C}{2})=tan(\cfrac{\Pi}{2}−\cfrac{A+B}{2})=\cfrac{1}{tan(\cfrac{A+B}{2})}=\cfrac{1-tan(\cfrac{A}{2})tan(\cfrac{B}{2})}{tan(\cfrac{A}{2})+tan(\cfrac{B}{2})}$$
Let $$a=tan(\cfrac{A}{2}), b=tan(\cfrac{B}{2}),c=tan(\cfrac{C}{2})$$, all positive, the constraint becomes
$$c=\cfrac{(1-ab)}{(a+b)}$$
which is equivalent to $$ab+bc+ca=1$$ 
In general, for any $$a,b,c$$
$$(a-b)^2 + (b-c)^2+(c-a)^2 \geq 0$$
$$2(a^2+b^2+c^2) \geq 2(ab+bc+ca)$$
$$\implies tan^2(\cfrac{A}{2}) + tan^2(\cfrac{B}{2}) + tan^2(\cfrac{C}{2}) \geq 1$$