Find the sup of this set with condition $x^2+y^2+z^2=1$ let $x,y,z&gt;0$,and such $x^2+y^2+z^2=1$,show that  $$\min{((x-y)^2,(y-z)^2,(z-x)^2)}&lt; \dfrac{1}{5}$$and show that $\dfrac{1}{5}$ is best constant. If it is not possible to find the maximal value, is it possible to give some reasonable upper bound?
Say $x\le y \le z$. So we have to show that at least one of the differences $y-x$, $z-y$ is  $&lt;\frac{1}{\sqrt{5}}$. Assume not. Then 
$x&gt;0$, $y &gt;\frac{1}{\sqrt{5}}$, $z &gt; \frac{2}{\sqrt{5}}$, hence their sum of squares is larger than $1$. One can easily see that the constant $1/5$ cannot be decreased, by using $(x,y,z)$ close to the triplet $(0, \frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$.
The best upper bound is $1/5$, and is achieved at six points on the sphere. These points have coordinates being the six permutations of $2/\sqrt{5},1/\sqrt{5},0$

To solve this, notice that given a point $P(x_0, y_0, z_0)$, $(x_0-y_0)^2$ is the square of the distance of point $P$ to the plane $x-y=0$, and similarly for $(y_0-z_0)^2$, $(z_0-x_0)^2$. So to maximize $\min\{(x_0-y_0)^2 ,(y_0-z_0)^2,(z_0-x_0)^2\}$, you should find a point on the sphere that's simultaneously far away from all these three planes.

If you imagine looking down at the sphere from the point $(1,1,1)$, you'll see the three planes cutting the sphere at $60$ degrees angle between them. Then it's clear from geometry which points should maximize $\min\{(x_0-y_0)^2 ,(y_0-z_0)^2,(z_0-x_0)^2\}$