Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is 
If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$
Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is

Try:
From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$
$$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$
$$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$
and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$
Could some help me how to find its upper bound, Thanks

Since $$a_{1}+a_{2}+a_{3}+a_{4}=0,$$ we can denote $$a_1+a_3=-(a_2+a_4)=t.$$Thus,
\begin{align}
&(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2\\=&2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})-2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})\\
=&2-2(a_{1}+a_{3})(a_{2}+a_{4})\\
=&2+2t^2.
\end{align}
Notice that
$$a_1a_3 \leq \frac{1}{4}(a_1+a_3)^2,$$ and 
$$ a_2a_4 \leq \frac{1}{4}(a_2+a_4)^2$$
Hence,
$$2(a_1a_3+a_2a_4) \leq t^2.$$
But
$$2(a_1a_3+a_2a_4)=(a_1+a_3)^2+(a_2+a_4)^2-(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})=2t^2-1.$$
Therefore, $$2t^2-1 \leq t^2.$$Thus,
$$t^2 \leq 1.$$
As a result,$$2+2t^2 \leq 2+2 \cdot 1=4,$$with equality holding if and only if $a_1=a_3,a_2=a_4$.
