MHB Maximize (x1+x2)(x1+x3)x4 for Quartic Equation with Real Roots in [1/2,2]

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The discussion focuses on maximizing the expression (x1+x2)(x1+x3)x4/(x4+x2)(x4+x3)x1 for a quartic equation with real roots constrained between 1/2 and 2. Observations indicate that the maximum occurs when the largest root, x4, is less than or equal to the other roots, while the minimum occurs in the opposite arrangement. The coefficients of the quartic equation are linked to the roots through specific relationships, leading to the conclusion that for a case where the product of the roots, d, equals 1, the maximum value of the expression is consistently 1. The analysis suggests that reciprocal pairs among the roots maintain this maximum value. The findings emphasize the symmetry and constraints of the roots in determining the expression's extremum.
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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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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]
 
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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
x_4 \leq x_i \leq x_1
and it is minimum, the reciprocal of the maximum, when
x_1 \leq x_i \leq x_4

2
Looking at the coefficients we get
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)
d=x_1x_2x_3x_4...(2)
b=x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4

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

(1) is written as
(x_1+x_4)(x_2x_3-1)+(x_1x_4-1)(x_2+x_3)=0
(x_2+A)(x_3+A)=1+A^2
where
A=\frac{x_1x_4-1}{x_1+x_4}
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.
 
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