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sum a^3b^3/(a+b) >=1/2 (1/a+1/b+1/c) for a,,c>0 with abc=1

Source: Mathcenter Contest / Oly - Thai Forum 2011 (R1) sl-5 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

November 15, 2022
algebrainequalities

Problem Statement

Let a,b,c∈R+a,b,c\in R^+ with abc=1abc=1. Prove that a3b3a+b+b3c3b+c+c3c3c+a≥12(1a+1b+1c)\frac{a^3b^3}{a+b}+\frac{b^3c^3}{b+c}+\frac{c^3c^3}{c+a} \ge \frac12 \left(\frac{1}{a}+ \frac{1}{b}+\frac{1}{c}\right)
(Zhuge Liang)
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