angles of a triangle

Posts: 791
Joined: Aug 4, 2011

by hemangsarkar, Sep 7, 2012, 7:04 AM

Q) if $A,B,C$ are the angles of a triangle, prove that -

$(\sin(A) + \sin(B) + \sin(C))^2 \le 6(1+\cos(A)\cos(B)\cos(C))$

my solution :

tried something new with this problem. jensen's inequality!!

$x^2$ is a convex function since its second derivative is always positive.

so, we must have

$\left(\frac{\sin(A) + \sin(B) + \sin(C)}{3} \right)^2 \le \frac{\sin^2(A) + \sin^2(B) + \sin^2(C)}{3}$

equality holds iff the triangle is equilateral.

also, $\sin^2(A) + \sin^2(B) + \sin^2(C) = 2 + \sin^2(A) - \cos^2(B) - \cos^2(C)$

$\cos^2(B) - \sin^2(A) = \cos(B+A)\cos(B-A)$
and $\cos(B+A) = -\cos(C)$

combining all these and using the sum to product formulas, we get

$\sin^2(A) + \sin^2(B) + \sin^2(C) = 2(1 + \cos(A)\cos(B)\cos(C))$

and the result follows.

Comment