The reals w, x, y, z all lie between -π/2 and π/2 and satisfy sin w + sin x + sin y + sin z = 1, cos 2w + cos 2x + cos 2y + cos 2z ≥ 10/3. Prove that they are all non-negative and at most π/6.
Solution
cos 2k = 1 - 2 sin2k. So put a = sin w, b = sin x, c = sin y, d = sin z and we have a + b + c + d = 1, a2 + b2 + c2 + d2 ≤ 1/3. Hence (a - 1/6)2 + (b - 1/6)2 + (c - 1/6)2 + (d - 1/6)2 = a2 + b2 + c2 + d2 - (a + b + c + d)/3 + 1/9 <= 1/9. Hence |a|, |b|, |c|, |d| ≤ 1/6 + 1/3 = 1/2.
Suppose one of a, b, c, d is negative. wlog we may take a < 0. Then b + c + d > 1, b2 + c2 + d2 < 1/3. Hence (b - 1/3)2 + (c - 1/3)2 + (d - 1/3)2 = (b2 + c2 + d2) - 2(b + c + d)/3 + 1/3 < 0. Contradiction. So 0 ≤ a, b, c, d ≤ 1/2. Hence 0 ≤ w, x, y, z ≤ π/6.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002