(Maybe Very) Hard Geometric Inequality

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Corei13
Posts: 153
Joined: Tue Dec 07, 2010 9:10 pm
Location: Chittagong

(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
Location: 24.4333°N 90.7833°E

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.

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