A triangle is drawn in a region bounded by the $x$-axis and a polynomial curve with all real roots, with one side of the triangle on the $x$-axis, as shown in this example:

What is the supremum of the ratio of the triangle's area to the region's area?

(We require that the polynomial has all real roots, because otherwise the answer would be trivial: the supremum would be $1$, for example with $y=x-x^n$ and letting $n\to\infty$.)

I hypothesize that the supremum of the ratio is approached with a region bounded by the $x$-axis from $x=-1$ to $x=1$, and $y=(\pm)(x^2-1) \prod\limits_{k=1}^n \left(\left(\frac{x}{r_k}\right)^2-1\right), r_k\ge1$, $r_k$'s not necessarily unique.

Using this hypothesis, the largest ratio $R$ that I have found is

$R\approx0.894934$ with $y=(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{47}$

Strangely, if we adjust the power of the $\left(\left(\frac{x}{5}\right)^2-1\right)$ factor, the ratio has a maximum when the power is $47$.

$R\approx0.894915$ with $y=-(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{46}}$

$R\approx0.894934$ with $y=(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{47}}$

$R\approx0.894933$ with $y=-(x^2-1)\left(\left(\frac{x}{5}\right)^2-1\right)^{\color{red}{48}}$

And if we adjust the value of the repeated roots, the ratio has a maximum when the repeated root is about $5$.

$R\approx0.892010$ with $y=(x^2-1)\left(\left(\frac{x}{\color{red}{4}}\right)^2-1\right)^{47}$

$R\approx0.894934$ with $y=(x^2-1)\left(\left(\frac{x}{\color{red}{5}}\right)^2-1\right)^{47}$

$R\approx0.890975$ with $y=(x^2-1)\left(\left(\frac{x}{\color{red}{6}}\right)^2-1\right)^{47}$