In which interval lies the minimum? Let $a_1,a_2,a_3,a_4$ be real numbers such that $a_1+a_2+a_3+a_4 =0 $ and $a_1^2+a_2^2+a_3^2+a_4^2=1$.
Then in what interval does  the smallest possible value of the following expression lies?
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$$
Here lies the actual question with the options.


Let $a_1=0$, $a_2=\frac{1}{\sqrt2},$ $a_3=0$ and $a_4=-\frac{1}{\sqrt2}.$
Hence, we get a value $2$.
We'll prove that it's a minimal value.
Indeed, let $a_1=a$, $a_2=b$, $a_3=c$.
Thus, $a_4=-a-b-c$, $a^2+b^2+c^2+(a+b+c)^2=1$ and we need to prove that
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\geq2$$ or
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\geq2\left(a^2+b^2+c^2+(a+b+c)^2\right)$$ or
$$(a+c)^2\geq0.$$
Done!
We'll find a maximal value for the collection.
Let $a_i=(-1)^{i-1}\frac{1}{2}.$ 
Hence, we get a value $4$.
We'll prove that it's a maximal value.
Indeed, let $a_1=a$, $a_2=b$, $a_3=c$.
Thus, $a_4=-a-b-c$, $a^2+b^2+c^2+(a+b+c)^2=1$ and we need to prove that
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\leq4$$ or
$$(a-b)^2+(b-c)^2+(a+b+2c)^2+(2a+b+c)^2\leq4\left(a^2+b^2+c^2+(a+b+c)^2\right)$$ or $$(a+b)^2+(b+c)^2\geq0.$$
Done!
