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a+b+c+1/abc \ge 4\sqrt3 if a^2 +b^2 +c^2 = 1 with a,b,c >0

Source: North Macedonian Mathematical Olympiad 1999 p5

February 12, 2020
inequalitiesalgebra

Problem Statement

If a,b,ca,b,c are positive numbers with a2+b2+c2=1a^2 +b^2 +c^2 = 1, prove that a+b+c+1abc≥43a+b+c+\frac{1}{abc} \ge 4\sqrt3
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