Question

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) ${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}>1.$

Answer 1

$1.$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.)

$f^{{}^{\prime }}\left(x\right)=cosx,f^{{}^{\prime \prime }}\left(x\right)=-sinx<0$, so the function is concave and we we'll apply the Jensen's inequality, which says that:

$f\left[\frac{A+B+C}{3}\right]>\frac{f\left(A\right)+f\left(B\right)+f\left(C\right)}{3}$

Working in a triangle, $A+B+C={180}^{\circ }$ and $\frac{(A+B+C)}{3}=\frac{180}{3}={60}^{\circ }$

$f\left(60\right)=sin{60}^{\circ }=\frac{\sqrt{3}}{2}$

$\frac{f\left(A\right)+f\left(B\right)+f\left(C\right)}{3}=\frac{sinA+sinB+sinC}{3}$

Putting the results again in Jensen's inequality:

$\frac{sinA+sinB+sinC}{3}<\frac{\sqrt{3}}{2}$

$sinA+sinB+sinC<\frac{3\left(\sqrt{3}\right)}{2}q.e.d.$

$2.$ $tan\left(\frac{C}{2}\right)=tan(\frac{\Pi }{2}-\frac{A+B}{2})=\frac{1}{tan\left(\frac{A+B}{2}\right)}=\frac{1-tan\left(\frac{A}{2}\right)tan\left(\frac{B}{2}\right)}{tan\left(\frac{A}{2}\right)+tan\left(\frac{B}{2}\right)}$

Let $a=tan\left(\frac{A}{2}\right),b=tan\left(\frac{B}{2}\right),c=tan\left(\frac{C}{2}\right)$, all positive, the constraint becomes

$c=\frac{(1-ab)}{(a+b)}$

which is equivalent to $ab+bc+ca=1$

In general, for any $a,b,c$

$(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\ge 0$

$2(a^2+b^2+c^2)\ge 2(ab+bc+ca)$

$⟹ta{n}^2\left(\frac{A}{2}\right)+ta{n}^2\left(\frac{B}{2}\right)+ta{n}^2\left(\frac{C}{2}\right)\ge 1$