Mathematical Reflections 2013, Issue 2 - Problem J259

Problem:
Among all triples of real numbers $(x,y,z)$ which lie on a unit sphere $x^2+y^2+z^2=1$ find a triple which maximizes
$\min (|x-y|, |y-z|, |z-x|)$.

Proposed by Arkady Alt.

Solution:
Suppose without loss of generality that $\min (|x-y|, |y-z|, |z-x|)=|x-y|$. Let $f(x,y,z)=|x-y|$ and $g(x,y,z)=x^2+y^2+z^2-1$. Consider the Lagrangian function $$\begin{array}{lll} L(x,y,z,\lambda)&=&f(x,y,z)-\lambda g(x,y,z)\\&=&|x-y|- \lambda(x^2+y^2+z^2-1), \end{array}$$ with $\lambda \in \mathbb{R}$. By Lagrange Multipliers Theorem, a maximum or a minimum for $f(x,y,z)$ subject to the constraint $g(x,y,z)=0$ must be a stationary point of $L$. Therefore a maximum or a minimum satisfies
$$\begin{array}{rcl} \dfrac{\partial L}{\partial x} & = & 0 \\ \dfrac{\partial L}{\partial y} & = & 0 \\  \dfrac{\partial L}{\partial z} & = & 0 \\ \dfrac{\partial L}{\partial \lambda} & = & 0, \end{array}$$ i.e. $$\begin{array}{rcl} \pm 1 - 2\lambda x & = & 0 \\ \mp 1 - 2\lambda y & = & 0 \\ -2\lambda z & = & 0 \\ x^2+y^2+z^2-1 & = & 0. \end{array}$$ From the third equation we get $z=0$ since $\lambda=0$ would give a contradiction in the first two equations. From the first two equations we have $x=\pm 1/2\lambda, y=\mp 1/2\lambda$ and substituting these values into the fourth equation we get $\lambda=\pm \sqrt{2}/2$, so $x=\pm \sqrt{2}/2, y=\mp \sqrt{2}/2, z=0$ are two stationary points which satisfies the conditions. It's easy to see that these two triples maximize $f(x,y,z)$ since a minimum for $f$ subject to the constraint $g$ is $0$ (take $x=y=0, z=1)$. By symmetry we find that all triples which maximize $\min (|x-y|, |y-z|, |z-x|)$ are $$(\pm \sqrt{2}/2, \mp \sqrt{2}/2, 0), (\pm \sqrt{2}/2, 0, \mp \sqrt{2}/2), (0, \pm \sqrt{2}/2, \mp \sqrt{2}/2),$$ and $\max (\min (|x-y|, |y-z|, |z-x|))=\sqrt{2}$.