Demonstrate that in the triangle ABC sinA + sinB + sinC

 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 degrees, 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[(A + B  + C)/3] > [f(A) + f(B) + f(C)]/3

Working in a triangle, A + B + C = 180 and (A +  B  + C)/3 = 180/3 = 60

f(60) = sin 60 = (sqrt 3)/2

[f(A) + f(B) + f(C)]/3 = (sin A + sin B + sin C)/3

Putting the results again in Jensen's inequality:

(sin A + sin B + sin C)/3 < (sqrt 3)/2

sin A + sin B + sin C < 3(sqrt 3)/2 q.e.d.