MathDB
Inequality with abc=1

Source: Tuymaada 2012, Problem 7, Day 2, Seniors

July 21, 2012
inequalitiesinequalities proposed

Problem Statement

Prove that for any real numbers a,b,ca,b,c satisfying abc=1abc = 1 the following inequality holds 12a2+b2+3+12b2+c2+3+12c2+a2+3≤12.\dfrac{1} {2a^2+b^2+3}+\dfrac {1} {2b^2+c^2+3}+\dfrac{1} {2c^2+a^2+3}\leq \dfrac {1} {2}.
Proposed by V. Aksenov
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