Problem Of The Week #495 February 6th 2022

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  • #1
anemone
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MHB
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Here is this week's POTW:

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Assume that the quartic equation $x^4-ax^3+bx^2-ax+d=0$ has four real roots $x_1,\,x_2,\,x_3,\,x_4$ where $\dfrac{1}{2}\le x_1,\,x_2,\,x_3,\,x_4 \le 2$. Find the maximum possible value of $\dfrac{(x_1+x_2)(x_1+x_3)x_4}{(x_4+x_2)(x_4+x_3)x_1}$.

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Answers and Replies

  • #2
bob012345
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Here is my answer.


My answer is 1.5625

Consider the expression as a function;
$$f(x_1,x_2,x_3,x_4) = \dfrac{(x_1+x_2)(x_1+x_3)x_4}{(x_4+x_2)(x_4+x_3)x_1}$$ we can take the partial derivatives to get maximize and minimum conditions.

$$ \frac{\partial f}{\partial x_1} = \frac{x_4(x_1^2 - x_2x_3)}{x_1^2(x_4 + x_2)(x_4 + x_3)} = 0$$
$$ \frac{\partial f}{\partial x_2} = \frac{x_4(x_1 + x_3)(-x_1 + x_4)}{x_1 (x_4 + x_3)(x_4 + x_2)^2} = 0$$
$$ \frac{\partial f}{\partial x_3} = \frac{x_4(x_1 + x_2)(-x_1 + x_4)}{x_1 (x_4 + x_2)(x_4 + x_3)^2} = 0$$
$$ \frac{\partial f}{\partial x_4} = \frac{(x_1 + x_2)(x_1 + x_3)(-x_4^2 + x_2x_3)}{x_1(x_4 + x_2)(x_4 + x_3)^2} = 0$$

Since all the roots are real and positive we can disregard relations that involve one root being the negative of the other leaving the following ##x_1 = x_4##, ##x_1 = \sqrt{x_2x_3}## and ##x_4 = \sqrt{x_2x_3}##

if ##x_1=x_4## then ##f(x_1,x_2,x_3,x_4)= 1## for all ##x_2,x_3##

There are many solutions to the second and third conditions but they suggest three roots are equal. Given that if we set ##x_1=x_2=x_3=\frac{1}{2}## and ##x_4=2## or vice versa we get a minimum of 0.64. If we set ##x_2=x_3=x_4=\frac{1}{2}## and ##x_1=2## or vice versa, we get a maximum of 1.5625. I also explored intermediate values all of which were lower.

If we put the condition for ##x_4## back into the original equation we get

$$f = \frac{(x_1 + x_4)^2}{4x_1x_4}$$ which maximizes to 1.5625 when ##x_1=\frac{1}{2}## and ##x_4=2## or vice versa. The minimum is one.

Also, if we put in the condition for ##x_1## back into the original equation we get

$$f = \frac{4x_1x_4}{(x_1 + x_4)^2}$$ which minimizes to 0.64 when ##x_1=\frac{1}{2}## and ##x_4=2## or vice versa. The maximum is one.

[\SPOILER]
 
Last edited:
  • #3
anuttarasammyak
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I have not solved the problem. The following is some observation.

1
It seems that the quantity, say ##I##, is maximum when large order of the solutions are
[tex]x_4 \leq x_i \leq x_1 [/tex]
and it is minimum, the reciprocal of the maximum, when
[tex]x_1 \leq x_i \leq x_4 [/tex]

2
Looking at the coefficients we get
[tex]a=x_1+x_2+x_3+x_4=\frac{d}{x_1}+\frac{d}{x_2}+\frac{d}{x_3}+\frac{d}{x_4}....(1)[/tex]
[tex]d=x_1x_2x_3x_4....(2)[/tex]
[tex]b=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4[/tex]

Thus for an easy case of d=1, in the maximum case
[tex]x_4=\beta^{-1},\ \ \{x_3,x_2\}=\{\alpha^{-1},\alpha\}, x_1=\beta[/tex]
where
[tex]1 \leq \alpha \leq \beta \leq 2[/tex]
then
[tex]I=1[/tex]
The minimum is also 1, so I=1 for any case.

(1) is written as
[tex](x_1+x_4)(x_2x_3-1)+(x_1x_4-1)(x_2+x_3)=0[/tex]
[tex](x_2+A)(x_3+A)=1+A^2[/tex]
where
[tex]A=\frac{x_1x_4-1}{x_1+x_4}[/tex]
Here we know that if there is a pair of reciprocal in the roots, the remaining pair is also reciprocal and d=1,I=1.
 
Last edited:

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