Geometric Inequality

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Geometric Inequality

Unread post by Nirjhor » Wed May 06, 2015 5:30 pm

Let $P$ be a point in the interior of $\triangle ABC$ with circumradius $R$. Prove that \[\dfrac{AP}{BC^2}+\dfrac{BP}{CA^2}+\dfrac{CP}{AB^2}\ge \dfrac 1 R.\]
Source: CleverMath homepage.

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- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

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Re: Geometric Inequality

Unread post by sowmitra » Wed May 20, 2015 12:51 am

Any hints? I'm stuck... :|
"Rhythm is mathematics of the sub-conscious."
Some-Angle Related Problems;

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