I have the following problem. I have the cubic polynomial $$x^3 + (2 - 2 n) x^2 + (2 - 3 n + n^2 + n s - s^2) x + n^2 - n^2s +n s^2 -n.$$
In this polynomial, $n\geq 4$ is even, $2\leq s\leq \frac{n}{2}$ and $s$ is an integer. I'd like to know for which of these values of $s$, the maximum root of the polynomial is maximized.
Since the root solutions of above polynomial are complex, I can't compare them directly.
Ps: I used Mathematica and tried some specific values. I found that when $s=n/2$, the maximum root is maximized. But I can't prove it. Any idea would be great for me.
A numerical study.
Assuming a real root $x_1$ and two complex conjugate $(x-b)^2+c^2$ we have
$$ x^3 + (2 - 2 n) x^2 + (2 - 3 n + n^2 + n s - s^2) x + n^2 - n^2s +n s^2 -n=(x-x_1)((x-b)^2+c^2) $$
and comparing coefficients we have
$$ \cases{ g_1 =x_1 \left(b^2+c^2\right)+n \left(s^2-n s+n-1\right)=0\\ g_2 = n^2+n (s-3)-b (b+2 x_1)-c^2-s^2+2=0\\ g_3 = 2 b-2n+x_1+2 = 0} $$
now we form the lagrangian
$$ L(x_1,s,b,c,\lambda) = x_1 + \lambda_1 g_1+ \lambda_2 g_2+ \lambda_3 g_3 $$
Solving for the stationary points and particularizing for $n = 4$
$$ \left( \begin{array}{cccc} x_1 & s & b & c \\ 0 & 1 & 3 & 0 \\ 0 & 3 & 3 & 0 \\ -2 \sqrt{3} & \frac{1}{2} \left(4-\frac{1}{2} \sqrt{160+96 \sqrt{3}}\right) & \frac{1}{4} \left(12+4 \sqrt{3}\right) & 0 \\ -2 \sqrt{3} & \frac{1}{2} \left(4+\frac{1}{2} \sqrt{160+96 \sqrt{3}}\right) & \frac{1}{4} \left(12+4 \sqrt{3}\right) & 0 \\ \end{array} \right) $$
for $n = 6$
$$ \left( \begin{array}{cccc} x_1 & s & b & c \\ 0 & 1 & 5 & 0 \\ 0 & 5 & 5 & 0 \\ \frac{1}{2} \left(2-2 \sqrt{21}\right) & \frac{1}{2} \left(6-\frac{1}{2} \sqrt{248+56 \sqrt{21}}\right) & \frac{1}{4} \left(18+2 \sqrt{21}\right) & 0 \\ \frac{1}{2} \left(2-2 \sqrt{21}\right) & \frac{1}{2} \left(6+\frac{1}{2} \sqrt{248+56 \sqrt{21}}\right) & \frac{1}{4} \left(18+2 \sqrt{21}\right) & 0 \end{array} \right) $$
etc,