Suppose $a$, $b$, and $c$ are the lengths of the sides of a triangle, and $R$ and $r$ are its circumradius and inradius respectively. How can one prove the following inequality? $$2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$$
Geometric inequality: $2r^2+8Rr \leq \frac{a^2+b^2+c^2}{2}$
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$\begingroup$
geometry
inequality
triangles
geometric-inequalities
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0I have modified the problem. In this way it should be easier to be proven. I don't have the solution. The image is irrelevant to the problem. I will delete it. – 2011-12-31
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8You, now, need to tell us what your symbols represent. – 2011-12-31
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1It is well known - R and r are the circumradius and inradius. a,b,c are the sides of the triangle. – 2011-12-31
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0If this is taken from a math competition, [tag:contest-math] tag would be appropriate. – 2011-12-31
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0Not really. Proved it using vectors but was looking for a simple proof. Well done. – 2011-12-31
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3@Chun-Yue Could you post your vector proof, please? I would like to see it. – 2011-12-31
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0Two suggestions were already made (to add the tag (contest-math) and to post your own proof), which you chose to ignore. So... let me add a third one: accept an answer (or explain why none satisfies you). – 2012-02-12