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

You can square the given equation:
$$a_{1}+a_{2}+a_{3}+a_{4}=0 \Rightarrow \\
a_1^2+a_2^2+a_3^2+a_4^2+2(a_1a_2+a_1a_3+a_1a_4+a_2a_3+a_2a_4+a_3a_4)=0 \Rightarrow \\
2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})=-1-2a_1a_3-2a_2a_4.$$
Hence:
$$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2=\\
=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})\color{red}{-}2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})=\\
2-(-1-2a_1a_3-2a_2a_4)=3+2a_1a_3+2a_2a_4\le 3+(a_1^2+a_3^2)+(a_2^2+a_4^2)=4.$$
the equality occurs for $a_1=a_3, a_2=a_4$.
