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sum a/ \sqrt{2b^2+2c^2-a^2} >= \sqrt3 for sidelengths a,b,c

Source: Mathcenter Contest / Oly - Thai Forum 2012 sl-4 https://artofproblemsolving.com/community/c3196914_mathcenter_contest

November 13, 2022
geometric inequalityinequalitiesalgebrageometry

Problem Statement

Let a,b,ca,b,c be the side lengths of any triangle. Prove that a2b2+2c2−a2+b2c2+2a2−b2+c2a2+2b2−c2≥3.\frac{a}{\sqrt{2b^2+2c^2-a^2}}+\frac{b}{\sqrt{2c^2+2a^2-b^2 }}+\frac{c}{\sqrt{2a^2+2b^2-c^2}}\ge \sqrt{3}. (Zhuge Liang)
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