Define f(x) = 1 - a cos x - b sin x - A cos 2x - B sin 2x, where a, b, A, B are real constants. Suppose that f(x) ≥ 0 for all real x. Prove that a2 + b2 ≤ 2 and A2 + B2 ≤ 1.
Solution
Take y so that cos y = a/√(a2 + b2), sin y = b/√(a2 + b2), and z so that cos 2z = A/√(A2 + B2), sin 2z = B/√(A2 + B2). Then f(x) = 1 - c cos(x - y) - C cos2(x - z), where c = √(a2 + b2), C = √(A2 + B2).
f(z) + f(π + z) ≥ 0 gives C ≤ 1. f(y + π/4) + f(y - π/4) ≥ 0 gives c ≤ √2.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
26 Jun 2001