Chemistry - Pedoe's inequality

In geometry, Pedoe's inequality, named after Dan Pedoe, states that if a, b, and c are the lengths of the sides of a triangle with area f, and A, B, and C are the lengths of the sides of a triangle with area F, then

$A^2(b^2+c^2-a^2)+B^2(a^2+c^2-b^2)+C^2(a^2+b^2-c^2)\geq 16Ff,\,$

with equality if and only if the two triangles are similar.

Note that the expression on the left is not only obviously symmetric in a, b, and c, and separately symmetric in A, B, and C, but also, perhaps slightly less obviously, remains the same if a is interchanged with A and b with B and c with C. In other words, it is a symmetric function of the pair of triangles.

References

"A Two-Triangle Inequality", D. Pedoe, The American Mathematical Monthly, volume 70, number 9, page 1012, November, 1963.

"An Inequality for Two Triangles", D. Pedoe, Proceedings of the Cambridge Philosophical Society, volume 38, part 4, page 397, 1943.

Categories: Inequalities