Problem 2 of Canadian Mathematical Olympiad 1969

Determine which of the two numbers \sqrt{c + 1} - \sqrt{c}, \sqrt{c} - \sqrt{c - 1} is greater for any c ≥ 1.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Since c ≥ 1, we always have c > \sqrt{c - 1}

or 2c > 2 \sqrt{c - 1}

or 4c > 2c + 2 \sqrt{c - 1} = [ \sqrt{c + 1}+ \sqrt{c - 1}]

or 2 \sqrt{c} > \sqrt{c + 1} + \sqrt{c - 1}

or \sqrt{c} - \sqrt{c - 1} > \sqrt{c + 1} - \sqrt{c}.