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!

This is a comment that’s too long to fit in the usual format. I prove below
the non-optimal bound $A\leq \sqrt{\frac{5+\sqrt{5}}{10}} \approx 0.8$. Let us denote
$$
x_1=x_6=a, x_2=b, x_3=c, x_4=d, x_5=e
$$
Then, we have
$$
\big(\frac{5+\sqrt{5}}{2}\big)
\bigg(\sum_{k=1}^5 x_k^2\bigg)-
\bigg(\sum_{k=1}^5 (x_k-x_{k+1})^2\bigg)=
\big(\frac{\sqrt{5}+1}{2}\big)
\bigg(x_5+\frac{(\sqrt{5}-1)(x_1+x_4)}{2}\bigg)^2+
\bigg(x_3+x_4-\frac{(\sqrt{5}-1)x_1}{2}\bigg)^2+
\big(\frac{\sqrt{5}-1}{2}\big)
\bigg(x_1+x_3+\frac{(\sqrt{5}+1)x_2}{2}\bigg)^2
$$
We deduce that
$$
\big(\frac{5+\sqrt{5}}{2}\big)
\bigg(\sum_{k=1}^5 x_k^2\bigg) \geq 
\bigg(\sum_{k=1}^5 (x_k-x_{k+1})^2\bigg)
$$
So if we put $\varepsilon={\sf min}(|x_{k+1}-x_k|)$, we have
$$
\frac{5+\sqrt{5}}{2} \geq \sum_{k=1}^5 (x_k-x_{k+1})^2 \geq 5\varepsilon^2
$$
and hence
$$
\varepsilon \leq \sqrt{\frac{5+\sqrt{5}}{10}}
$$
