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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?


You can find a proof in Topics in Inequalities - Theorems and Techniques by Hojoo Lee. See e.g. here, Theorem 1.3.2.

  • $\begingroup$ Oh, thanks! ${}$ $\endgroup$ – Sawarnik Mar 30 '14 at 20:20

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