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R \geq 2r, A little generalised

Source: India tst 2006 p7

June 26, 2012
geometryinradiuscircumcircleinequalitiesinequalities proposed

Problem Statement

Let ABCABC be a triangle with inradius rr, circumradius RR, and with sides a=BC,b=CA,c=ABa=BC,b=CA,c=AB. Prove that R2r≥(64a2b2c2(4a2−(b−c)2)(4b2−(c−a)2)(4c2−(a−b)2))2.\frac{R}{2r} \ge \left(\frac{64a^2b^2c^2}{(4a^2-(b-c)^2)(4b^2-(c-a)^2)(4c^2-(a-b)^2)}\right)^2.
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