# Extrema of x^3+y^3+z^3-xyz on the unit sphere

Grégoire Nicollier
February 11, 2016

See http://www.cut-the-knot.org/Optimization/MD.shtml for another proof.

x^3+y^3+z^3-xyz can be factored as (x+y+z)(x^2+y^2+z^2-xy-yz-zx). We thus have to optimize

(*)    (x+y+z)(1-xy-yz-zx) on the unit sphere.

Consider the unit vector \vec v=(x,y,z) forming the angle \alpha with the axis \vec a=(1,1,1). Turn \vec v about 120^\circ around \vec a to get \vec v_t=(y,z,x). It is easy to see that angle \beta between \vec v and \vec v_t fulfills \displaystyle\sin\frac\beta2=\frac{\sqrt3}2\sin\alpha and thus

\displaystyle1-\cos\beta=2\sin^2\frac\beta2=\frac32\sin^2\alpha.

Considering dot products, (*) is nothing but

\displaystyle\sqrt3\cos\alpha\,(1-\cos\beta)=\frac{3\sqrt3}2\cos\alpha\sin^2\alpha=\frac{3\sqrt3}2(u-u^3) for u=\cos\alpha.

The odd function u-u^3 is extremal at \cos\alpha=u=\pm 1/\sqrt3, i.e., (*) is constant on every circle of the unit sphere that is perpendicular to \vec a, (*) is extremal on the circles lying in the planes x+y+z=\pm1, and (*) reaches there the extremal values \pm1.