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Easy inequality with ab+bc+ca=1

Source: Iran TST 2012-Second exam-2nd day-P4

May 13, 2012
inequalitiesinequalities proposed

Problem Statement

For positive reals a,ba,b and cc with ab+bc+ca=1ab+bc+ca=1, show that 3(a+b+c)≤aabc+bbca+ccab.\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})\le \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}.}
Proposed by Morteza Saghafian
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