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 $a,b,c$
$(a-b)^2 + (b-c)^2+(c-a)^2 \geq 0$
$2(a^2+b^2+c^2) \geq 2(ab+bc+ca)$
$\implies tan^2(\cfrac{A}{2}) + tan^2(\cfrac{B}{2}) + tan^2(\cfrac{C}{2}) \geq 1$