74. Inegalitatea Erdos-Mordell.

by Virgil Nicula, Aug 2, 2010, 3:56 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=136745&ml=1

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6073
Quote:
Let $D$ , $E$ , $F$ be projections of interior point $P$ on sidelines of $\triangle ABC$.

Then $\boxed {\ PA+PB+PC\ge 2\cdot (PD+PE+PF)\ }$ .

Proof. $PA\cdot\sin A=EF\ge \mathrm{Pr}_{BC}[EF]=$ $\mathrm{Pr}_{BC}[PE]+\mathrm{Pr}_{BC}[PF]=$ $PE\cdot \sin C+PF\cdot\sin B$ $\implies$ $\boxed {a\cdot PA\ge c\cdot PE+b\cdot PF}$ . Remark that if consider symmetrical point of $P$ w.r.t. bisector of $\widehat{BAC}$ and apply same method then we"ll obtain $\boxed {a\cdot PA\ge b\cdot PE+c\cdot PF}$ . Obtain analogously $b\cdot PB\ge a\cdot PF+c\cdot PD$ and $c\cdot PC\ge b\cdot PD+a\cdot PE$ . Therefore, $\sum PA\ge \sum\left(\frac bc+\frac cb\right)\cdot PD\ge 2\cdot\sum PD$ .

Applications.

$1\blacktriangleright\ \ \frac{PA}{h_b+h_c} + \frac{PB}{h_a+h_c} + \frac{PC}{h_a+h_b} \ge 1$ .

$2\blacktriangleright\ \ \frac {PA}{a^{2}} + \frac {PB}{b^{2}} + \frac {PC}{c^{2}}\ge \frac 1R$ .
This post has been edited 8 times. Last edited by Virgil Nicula, Nov 23, 2015, 3:32 PM

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