Exercise 1321Geometry › Plane geometryOlympiadAdded 5 Nov 2026

Weitzenböck's inequality

#geometric inequality#Heron's formula#area#sum of squares

Statement

In a triangle with sides $a,b,c$ and area $S$ , prove that
$a^2+b^2+c^2\ge4\sqrt3\,S,$
with equality iff the triangle is equilateral.

Hint

Use Heron's formula in the form $16S^2=2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$ , and square the inequality.

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Heron's formula. $16S^2=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)$ .

Reduction. Both sides of the claim are positive, so it is equivalent to its square, which becomes
$a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2.$

Proof of the last inequality.
$a^4+b^4+c^4-a^2b^2-b^2c^2-c^2a^2=\tfrac12\left[(a^2-b^2)^2+(b^2-c^2)^2+(c^2-a^2)^2\right]\ge0.$

Equality. It needs $a^2=b^2=c^2$ , i.e. $a=b=c$ .

Remark. The constant $4\sqrt3$ is sharp. Hadwiger–Finsler strengthens this to $a^2+b^2+c^2\ge4\sqrt3S+(a-b)^2+(b-c)^2+(c-a)^2$ .

Lesson. A geometric inequality involving an area becomes algebraic through Heron's formula; after squaring it is often a sum of squares.