## (Maybe Very) Hard Geometric Inequality

For discussing Olympiad level Geometry Problems
Corei13
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### (Maybe Very) Hard Geometric Inequality

Let $\triangle ABC$ is a triangle with circumradius $R$. Extend $AB$ to $R$, $BC$ to $P$ an $CA$ to $Q$ and let $D,E,F$ be the midpoints of $BC$, $CA$, $AB$ respectively.
Prove that,
$\frac{PQ}{AB}+\frac{QR}{BC}+\frac{RP}{CA} \ge \frac{PD+QE+RF}{R}$
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sakibtanvir
Posts: 188
Joined: Mon Jan 09, 2012 6:52 pm
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### Re: (Maybe Very) Hard Geometric Inequality

Corei13 wrote:Let $\triangle ABC$ is a triangle with circumradius $R$. Extend $AB$ to $R$, $BC$ to $P$ an $CA$ to $Q$ and let $D,E,F$ be the midpoints of $BC$, $CA$, $AB$ respectively.
Prove that,
$\frac{PQ}{AB}+\frac{QR}{BC}+\frac{RP}{CA} \ge \frac{PD+QE+RF}{R}$ An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.

Phlembac Adib Hasan
Posts: 1016
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### Re: (Maybe Very) Hard Geometric Inequality

I think one $R$ is length and the other is a point.
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