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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?

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  • $\begingroup$ I don't see how you get the expresion <1. From what I see, you only come back where you started expression>0. $\endgroup$ – Julian Mejia Apr 25 at 4:55
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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 $2$.

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a1*a2 and a2*a3 and a3*a4 and a4*a1 all ranges from 0 to 1/2, but the trick here is not all of them can maximize to 1 at a single time, which takes down the maximum range further below, and now i have given you the pointer to solve it, you should try and revert back if still faced problem solving it

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