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6

Part of 2013 Czech And Slovak Olympiad IIIA

Problems(1)

\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

2/17/2020
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.
inequalitiesalgebra