Weizenbock’s inequality


In a triangleMathworldPlanetmath with sides a, b, c, and with area A, the following inequality holds:

a2+b2+c24A3

The proof goes like this: if s=a+b+c2 is the semiperimeter of the triangle, then from Heron’s formulaMathworldPlanetmathPlanetmath we have:

A=s(s-a)(s-b)(s-c)

But by squaring the latter and expanding the parentheses we obtain:

16A2=2(a2b2+a2c2+b2c2)-(a4+b4+c4)

Thus, we only have to prove that:

(a2+b2+c2)23[2(a2b2+a2c2+b2c2)-(a4+b4+c4)]

or equivalently:

4(a4+b4+c4)4(a2b2+a2c2+b2c2)

which is trivially equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath to:

(a2-b2)2+(a2-c2)2+(b2-c2)20

Equality is achieved if and only if a=b=c (i.e. when the triangle is equilateral) .

See also the Hadwiger-Finsler inequality, from which this result follows as a corollary.

Title Weizenbock’s inequality
Canonical name WeizenbocksInequality
Date of creation 2013-03-22 13:19:33
Last modified on 2013-03-22 13:19:33
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 9
Author mathcam (2727)
Entry type Theorem
Classification msc 51F99
Related topic HadwigerFinslerInequality