How find this inequality $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$ let $a,b,c,d,e\in R$,and such
$$a^2+b^2+c^2+d^2+e^2=1$$
 find this value
$$A=\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$$
I use computer have this 
$$A=\dfrac{2}{\sqrt{10}}$$
then equal holds if we suppose that $a\leq b\leq c\leq d\leq e$, then$$a=\frac{2}{\sqrt{10}},b=-\frac{1}{\sqrt{10}},c=\dfrac{1}{\sqrt{10}},d=-\frac{2}{\sqrt{10}},e=0$$
But I  consider sometimes, I want use Cauchy-Schwarz inequality to solve it,
I guess have this follow?
$$(|a-b|+|b-c|+|c-d|+|d-e|+|e-a|)^2\le(\dfrac{1}{10})(a^2+b^2+c^2+d^2+e^2)$$
then I  can't. Thank you very much!

You cannot suppose (without loss of generality) that $a\le b\le c\le d\le e$, because the quantity you're maximizing is not fully symmetric in the five variables. For example, taking $a=c=-\frac12$, $b=d=\frac12$, and $e=0$ shows that $A\ge\frac12$.
