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Barrow's inequality states that if $P$ is any point inside triangle $ABC$, and $PU$, $PW$, and $PV$ are the angle bisectors, then the following inequality holds, $PA+PB+PC\geq 2(PU+PV+PW)$.

I know that it is a stronger version of the Erdos-Mordell inequality, of which some very short and elegant proofs have been found. However, I have not been able to find any proof of the Barrow's inequality on the internet. Can anyone give me the proof?

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up vote 2 down vote accepted

I immediately found a proof here http://www.scribd.com/doc/25055669/Hojoo-Lee-Topics-in-Inequalities-Methods-and-Techniques on top of page 10 (and bottom of 9).

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Oh, thanks! ${}$ –  Sawarnik Mar 30 at 20:20

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