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

By Cauchy inequality we have: 
$$(a_1+a_3)^2 \leq 2(a_1^2+a_3^2)$$
and
$$(a_2+a_4)^2 \leq 2(a_2^2+a_4^2)$$
Since $(a_1+a_3)^2=(a_2+a_4)^2$ we have $$2(a_1+a_3)^2 \leq 2(a_1^2+a_3^2) +2(a_2^2+a_4^2) =2$$
So $$ E \leq  4$$
