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\sqrt{a^2 + pb^2} +\sqrt{b^2 + pa^2} \ge a + b + (p - 1) \sqrt{ab}

Source: Czech And Slovak Mathematical Olympiad, Round III, Category A 2013 p6

February 17, 2020
inequalitiesalgebra

Problem Statement

Find all positive real numbers pp such that a2+pb2+b2+pa2≥a+b+(p−1)ab\sqrt{a^2 + pb^2} +\sqrt{b^2 + pa^2} \ge a + b + (p - 1) \sqrt{ab} holds for any pair of positive real numbers a,ba, b.
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