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sum a/ \sqrt{b+c} >= (a+b+c) /\sqrt{2} if a,b,c>0, with a+b+c+abc=4

Source: Mathcenter Contest / Oly - Thai Forum 2012 sl-5 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

November 13, 2022
algebrainequalitiesHi

Problem Statement

Let a,b,c>0 and a+b+c+abc=4a+b+c+abc=4. Prove that ab+c+bc+a+ca+b≥12(a+b+c).\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b} }\ge \frac{1}{\sqrt{2}}(a+b+c).
(Zhuge Liang)
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