# 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:

1. $$(0,1.5)$$
2. $$(1.5,2.5)$$
3. $$(2.5,3)$$
4. $$(3,3.5)$$

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?

• I don't see how you get the expresion <1. From what I see, you only come back where you started expression>0. – Julian Mejia Apr 25 at 4:55

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 $$2$$.