Let y = |√(1 + sin 2x) - √(1 - sin 2x)|. Then y 2 = 2 - 2|cos 2x|. If x belongs to [0, π/4] or [3π/4, 5π/4] or [7π/4], then cos 2x is non-negative, so y 2 = 2 - 2 cos 2x = 4 sin 2 x, so y = 2|sin x|. We have cos x <= |sin x| except for x in [0, π/4] and [7π/4, 2π]. So that leaves [3π/4, 5π/4] in which we certainly have |sin x| ≤ 1/√2. If x belongs (π/4, 3π/4) or (5π/5, 7π/4), then cos 2x is negative, so y 2 = 2 + 2 cos 2x = 4 cos 2 x. So y = 2 |cos x|. So the first inequality certainly holds. The second also holds. Thus the inequalities hold for all x in [π/4, 7π/4]. s 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. 7th IMO 1965 © John Scholes jscholes@kalva.demon.co.uk 25 Sep 1998 Last corrected/updated 26 Sep 2003
IMO 1965 Problem A1 Find all x in the interval [0, 2π] which satisfy: 2 cos x ≤ |√(1 + sin 2x) - √(1 - sin 2x)| ≤ √2.