Barrow's inequality

Let P be an arbitrary point inside the triangle ABC. From P and ABC, define U, V, and W as the points where the angle bisectors of BPC, CPA, and APB intersect the sides BC, CA, AB, respectively. Then Barrow's inequality states that^[1]

${\displaystyle PA+PB+PC\geq 2(PU+PV+PW),\,}$

with equality holding only in the case of an equilateral triangle and P is the center of the triangle.^[1]

Barrow's inequality can be extended to convex polygons. For a convex polygon with vertices ${\displaystyle A_{1},A_{2},\ldots ,A_{n}}$ let ${\displaystyle P}$ be an inner point and ${\displaystyle Q_{1},Q_{2},\ldots ,Q_{n}}$ the intersections of the angle bisectors of ${\displaystyle \angle A_{1}PA_{2},\ldots ,\angle A_{n-1}PA_{n},\angle A_{n}PA_{1}}$ with the associated polygon sides ${\displaystyle A_{1}A_{2},\ldots ,A_{n-1}A_{n},A_{n}A_{1}}$, then the following inequality holds:^[2][3]

${\displaystyle \sum _{k=1}^{n}|PA_{k}|\geq \sec \left({\frac {\pi }{n}}\right)\sum _{k=1}^{n}|PQ_{k}|}$

Here ${\displaystyle \sec(x)}$ denotes the secant function. For the triangle case ${\displaystyle n=3}$ the inequality becomes Barrow's inequality due to ${\displaystyle \sec \left({\tfrac {\pi }{3}}\right)=2}$.

Barrow's inequality strengthens the Erdős–Mordell inequality, which has identical form except with PU, PV, and PW replaced by the three distances of P from the triangle's sides. It is named after David Francis Barrow. Barrow's proof of this inequality was published in 1937, as his solution to a problem posed in the American Mathematical Monthly of proving the Erdős–Mordell inequality.^[1] This result was named "Barrow's inequality" as early as 1961.^[4]

A simpler proof was later given by Louis J. Mordell.^[5]

• Euler's theorem in geometry
• List of triangle inequalities