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!
 A: 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 $A\ge\frac12$.
A: 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
$$
\varepsilon \leq \sqrt{\frac{5+\sqrt{5}}{10}}
$$
A: The situation described by the OP (situation (c) in the following figure) is optimal. This can be seen as follows:
We consider the dual problem instead: Five telescopic rods of  length $\geq1$ each are hinged together at the ends so that a flexible and extensible pentagon is formed. This pentagon is then extended and squeezed to a "linear structure" $S$, i.e., such that the five hinges are in line. This structure is then positioned along the $x$-axis. Let $a_i$ $\>(1\leq i\leq 5)$ be the resulting positions of the five hinges. Our aim is to minimize the quantity $$\Phi:=\sum_{i=1}^5a_i^2\ .$$
A first step in this direction is to translate $S$ along the $x$-axis such that the centroid of the hinges is at the origin. The resulting $\Phi$-value will be denoted by $\Phi'(S)$ and can be considered as total moment of inertia of the five hinges with respect to  their centroid $c=0$.
The "linear structure" $S$ has $2$ or $4$ return-hinges and the rest $180^\circ$-hinges. When there are $2$ return-hinges they can have $1$ or $2$ rods between them. These are the configurations (a) and (b) in the following figure. When there are $4$ return-hinges there is just one $180^\circ$-hinge, see (c) in the figure.

The red rods are critical: we have to ensure that their length is $\geq1$, whereas the black rods automatically have a length $>1$. The configurations (a)–(c) with all red rods of length 1 are possible and admissible. On the other hand, our physical intuition tells us that making any of these red rods longer than $1$ will increase the moment of inertia $\Phi'(S)$ of the configuration. In cases (b) and (c) some "internal shifting" is possible, but we know from experience that the symmetric situation has smallest moment of inertia. (I shall treat case (c) explicitly at the end.) It follows that the optimal positions in the three cases are given by
$${\rm (a)}\qquad a_1=-2,\quad a_2=-1, \quad a_3=0, \quad a_4=1,\quad  a_5=2,\qquad \Phi'=10\ ;$$
$${\rm (b)}\qquad a_1=-{3\over2},\quad a_2=-{1\over2}, \quad a_3={1\over2}, \quad a_4={3\over2},\quad  a_5=0,\qquad \Phi'=5\ ;$$
$${\rm (c)}\qquad a_1=-1,\quad a_2=0, \quad a_3=1, \quad a_4=-{1\over2},\quad  a_5={1\over2},\qquad \Phi'={5\over2}\ .$$
From this we conclude that the global minimum of $\Phi$, resp. $\Phi'$, is assumed in situation (c). 
Returning to the original problem we therefore can say the following: When $\Phi=1$ is prescribed then at least one of the rods must have a length $$d\leq\sqrt{2\over 5}\ .$$
Update: There was some handwaving above. In the following I shall treat case (c) in detail, the other cases are simpler. 
When the lengths $d_1$, $d_2$, $d_3$ of the three red rods are given then for suitable $u$, $v$ one has
$$a_1=v-{2d_1+d_2\over3}, \quad a_2=v+{d_1-d_2\over 3},\quad a_3=v+{d_1+2d_2\over3},$$ $$ a_4=u-{d_3\over2},\quad a_5=u+{d_3\over2}\ .$$
From this one computes
$$\Phi=3v^2+{2\over3}(d_1^2+d_1d_2+d_2^2)+2u^2+{1\over2}d_3^2\ .$$
It follows that for given $d_i$ $\>(1\leq i\leq 3)$ one may attain
$$\Phi_{\min}={2\over3}(d_1^2+d_1d_2+d_2^2)+{1\over2}d_3^2\ ,$$
which is an increasing function of the $d_i$, as claimed in the main text.
