Show that for any triangle, 3(√3)/2 ≥ sin 3A + sin 3B + sin 3C ≥ -2. When does equality hold?
Solution
Answer: 1st inequality, A = B = 20o, C = 140o. 2nd inequality, A = 0o, B = C = 90o (which is degenerate).
For x between 0 and 180o, sin 3x is negative iff 60o < x < 120o. So at most two angles x in a triangle can have sin 3x negative. Obviously sin 3x ≥ -1, so sin 3A + sin 3B + sin 3C ≥ -2. We can only get equality in the degenerate case A = 0o, B = C = 90o (or with the angles relabeled).
We can certainly achieve 3(√3)/2 as shown above. But 3(√3)/2 > 2, so we must have all three sines positive. [If only two are positive, then their sum is at most 2.]. sin 3x is positive for x < 60o and x > 120o. So one angle must be > 120o. Assume, for definiteness, that A ≤ B < 60o, C > 120o. Put C' = C - 120o. Then 3A + 3B + 3C' = 180o. Now sin x is a concave function for 0 ≤ x ≤ 180o, or - sin x is a convex function. So by Jensen's inequality -sin x - sin y - sin z ≥ 3 -sin(x + y + z)/3. Hence sin 3A + sin 3B + sin 3C ≤ 3 sin 60o = 3(√3)/2.
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002