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
 A: 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 $a_1=a_3,a_2=a_4$.
A: Let $\boldsymbol{x}=(a_1,a_2,a_3,a_4) \in \mathbb{R}^4$,
$$E=\boldsymbol{x}
\begin{pmatrix}
  2 & -1 & 0 & -1 \\
  -1 & 2 & -1 & 0 \\
  0 & -1 & 2 & -1 \\
  -1 & 0 &-1 & 2
\end{pmatrix}
\boldsymbol{x}^T$$


*

*Eigenvalues: $$\lambda_1=4, \, \lambda_2=\lambda_3=2, \, \lambda_4=0$$

*Unit eigenvectors:
\begin{align}
  \boldsymbol{v}_1 &= \frac{(-1,1,-1,1)}{2} \\
  \boldsymbol{v}_2 &= \frac{(0,-1,0,1)}{\sqrt{2}} \\
  \boldsymbol{v}_3 &= \frac{(-1,0,1,0)}{\sqrt{2}} \\
  \boldsymbol{v}_4 &= \frac{(1,1,1,1)}{2} \\
  \boldsymbol{v}_i \cdot \boldsymbol{v}_j &= \delta_{ij}
\end{align}


*

*Note that $\boldsymbol{v}_k$ satisfies $a_1+a_2+a_3+a_4=0$ for $k=1$, $2$ or $3$ whereas $\boldsymbol{v}_4$ is normal to the hyperplane $a_1+a_2+a_3+a_4=0$.  Therefore,


$$\boldsymbol{x}=
\alpha \, \boldsymbol{v}_1+
\beta \, \boldsymbol{v}_2+
\gamma \, \boldsymbol{v}_3$$
$$\lambda_{2,3} \Vert \boldsymbol{x} \Vert^2
\le E \le
\lambda_{1} \Vert \boldsymbol{x} \Vert^2$$
$$2\le E  \le 4$$
A: 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 $$ E \leq  4$$
A: 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 $a_1=a_3, a_2=a_4$.
A: You can try also like this. Write $$a_1 =\cos \alpha \sin \beta$$
$$a_2 =\cos \alpha \cos \beta$$
$$a_3 =\sin \alpha \sin \beta$$
$$a_4 =\sin \alpha \cos \beta$$
for some $\alpha$ and $\beta$.
