Let $\alpha <0 <\beta$ be real numbers and consider the polynomial $f(x)=x(x-\alpha)(x-\beta).$ Let $S$ be the set of real numbers $s$ such that the polynomial $f(x)-s$ has three real roots. For each $s \in S$ let $p(s)$ be the product of the least and largest root of $f(x)-s$. Determine the least possible value of $p(s)$ when varying $s$ in $S$ and the corresponding values of $s$ at which the minimum is obtained.
This question is from a math Olympiad http://ciim.uan.edu.co/ciim-2018-pruebas. I've been trying to get and answer but I don't seem to get anywhere. I tried using Vieta relations to express the product of the roots in terms of s, $\alpha$ and $\beta$ to then minimize $p(s)$ but no significant progress has been made. I appreciate any help.
Let
$$g(x) = f(x) - s = x^3 -(\alpha+\beta)x^2+\alpha\beta x-s$$
Setting
$$\begin{align*} x_N & = \dfrac{\alpha+\beta}{3} \\ \\ \delta^2 &= x_N^2 -\dfrac{\alpha\beta}{3}\\ \\ y_N &= g(x_N) = f(x_N)-s \\ \\ h &= 2\delta^3 \\ \\ \theta(s) &= \dfrac{1}{3}\arccos\left(\dfrac{-y_N}{h}\right) = \dfrac{1}{3}\arccos\left(\dfrac{s-f(x_N)}{h}\right) \end{align*}$$
The product of the greatest and least root of $g(x)$, $p(s)$ can be expressed as
$$p(s)=\left[x_N +2\delta\cos\left(\theta(s)\right)\right]\left[x_N +2\delta\cos\left(\theta(s)+\dfrac{2\pi}{3}\right)\right]$$
Note that $\theta(s)$ is restricted to $\left[0, \dfrac{\pi}{3}\right]$.
If you can make the argument that $p(s)$ is least when the greatest and least roots of $g(x)$ are farthest apart, then you can make a geometric argument, using a circle of radius $2\delta$ and the triangle formed by the circle's center and the two roots, that the two roots are farthest apart when $\theta(s) = \dfrac{\pi}{6}$, because the projection of the hypotenuse onto the x-axis is maximized for that angle. See figure 2 on page 6 of this paper for why that projection matters for the placement of the roots on the x-axis.
If $\theta(s) = \dfrac{\pi}{6}$, then
$$\begin{align*} s_{min} &= f(x_N) = f\left(\dfrac{\alpha+\beta}{3}\right) \\ \\ &= x_N^3 - (\alpha+\beta)x_N^2 +\alpha\beta x_N\\ \\ &= \left[\dfrac{\alpha+\beta}{3}\right]\left[-2\left(\dfrac{\alpha+\beta}{3}\right)^2+\alpha\beta\right] \\ \end{align*}$$
So
$$\begin{align*} p(s_{min}) &= \left(x_N + 2\delta\dfrac{\sqrt{3}}{2}\right)\left(x_N - 2\delta\dfrac{\sqrt{3}}{2}\right)\\ \\ &= x_N^2 -3\delta^2\\ \\ &= x_N^2 -3x_N^2 + \alpha\beta\\ \\ &= -2\left(\dfrac{\alpha+\beta}{3}\right)^2 + \alpha\beta\\ \end{align*}$$
And we end up with a curious relationship between the average of the roots, $x_N$, and $s_{min}$ and $p(s_{min})$:
$$s_{min} = x_N \cdot p\left(s_{min}\right)$$