IMO 1965 · Problem 1

Determine all values \(x\) in the interval \(0 \leq x \leq 2\pi\) which satisfy the inequality \[ 2\cos x \leq \left|\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}\right| \leq \sqrt{2}. \]

Official text of the problem. Official problem paper (PDF)

Answer

\[ \boxed{x\in\left[\frac{\pi}{4},\frac{7\pi}{4}\right].} \]

Solution

Let \[ D=\left|\sqrt{1+\sin 2x}-\sqrt{1-\sin 2x}\right|. \] Both square roots are defined for every real \(x\). Squaring gives \[ D^2 =2-2\sqrt{(1+\sin 2x)(1-\sin 2x)} =2-2\sqrt{\cos^2 2x} =2-2|\cos 2x|. \] In particular, \(0\leq D\leq\sqrt2\), so the upper inequality always holds.

If the given inequalities hold, then \[ 2\cos x\leq D\leq\sqrt2, \] and hence \(\cos x\leq 1/\sqrt2\). On \(0\leq x\leq2\pi\), this is equivalent to \[ \frac{\pi}{4}\leq x\leq\frac{7\pi}{4}. \]

Conversely, suppose \(x\) lies in this interval. If \(\cos x\leq0\), then \[ 2\cos x\leq0\leq D, \] so the lower inequality holds.

If \(\cos x>0\), then \(0<\cos x\leq1/\sqrt2\), and therefore \[ \cos 2x=2\cos^2x-1\leq0. \] Consequently, \[ D^2=2+2\cos 2x=4\cos^2x. \] Since \(D\geq0\) and \(\cos x>0\), this yields \(D=2\cos x\), so the lower inequality again holds.

Thus every point of the stated closed interval, including both endpoints, satisfies both inequalities.

The solution was checked against John Scholes' IMO solution archive (Kalva).