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abc = 2, a, b, c <= sqrt{2}

Source: Junior Olympiad of Malaysia Shortlist 2015 A3

July 17, 2015
Inequalityinequalities

Problem Statement

Let a,b,c a, b, c be positive real numbers less than or equal to 2 \sqrt{2} such that abc=2 abc = 2 , prove that 2∑cycab+3c3ab+c≥a+b+c \sqrt{2}\displaystyle\sum_{cyc}\frac{ab + 3c}{3ab + c} \ge a + b + c
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