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2019 Central American and Caribbean Mathematical Olympiad, P5

Source:

June 19, 2019
algebrainequalities

Problem Statement

Let a, ba,\ b and cc be positive real numbers so that a+b+c=1a+b+c=1. Show that aa2+6bc+bb2+6ac+cc2+6ab324a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}\leq\frac{3\sqrt{2}}{4}