The smallest value of the expression: $(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$ lies in which of the following intervals? Question:

Let $a_1,a_2,a_3,a_4\in\Bbb R$, 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 the smallest value of the expression, $$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2$$
lies in the interval:


*

*$(0,1.5)$

*$(1.5,2.5)$

*$(2.5,3)$

*$(3,3.5)$

My Attempt:

Here,
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2\gt0\qquad\forall a_i\in\Bbb R$$
Now:
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2 = 2(1)-2(a_1a_2+a_2a_3+a_3a_4+a_4a_1)$$
since,$$(a_1-a_2)^2\gt0\implies a_1^2+a_2^2 \gt 2a_1a_2$$
$$\implies 2(a_1^2+a_2^2+a_3^2+a_4^2)\gt 2(a_1a_2+a_2a_3+a_3a_4+a_4a_1)$$
$$\implies a_1a_2+a_2a_3+a_3a_4+a_4a_1\lt 1$$
Therefore,
$$(a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_4)^2+(a_4-a_1)^2\lt1$$
Therefore the answer (according to me) should be option $(1)$, or the interval: $(0,1.5)$,
however, the answer given is option $(2)$, or the interval: $(1.5,2.5)$.

Can someone tell me where my error is (or) a better method to solve this problem?

Hint: Your sum simplifies to $$2-2(a_1a_3+a_2a_3+a_3a_4+a_4a_1)=2-2(a_2(a_1+a_3)+a_4(a_1+a_3))=2-2(-(a_2+a_4)^2)$$ and $$2+2(a_2+a_4)^2\geq 2$$. The minimum is equal to $2$.
