Consider recursively defined polynomials $f_0(x) = x$ and $f_{n+1}(x) = f_n(x) - f_n'(x) x (1-x)$.
These polynomials have some special properties, for example $f_n(0) = 0$, $f_n(1) = 1$, and all $n+1$ roots of $f_n$ are in $[0,1)$. Let $x_n$ denote the largest root of $f_n$. Then $f_n(x_n) = 0$ and $f_n'(x_n)>0$. Moreover, $x_n > x_{n-1}$ for all $n$.
I want to prove the following claim: $f_{n}'(x_{n+1}) > f_{n-1}'(x_{n+1})$ for all $n \geq 2$.
Note that the claim does not hold for arbitrary $x$. The derivatives are polynomials themselves, by Gauss-Lucas theorem all their roots are in $[0,x_n)$ and there are many points where $f_{n}'(x) < 0 < f_{n-1}'(x)$. However, I am quite sure that at $x \geq x_{n+1}$, the derivatives are ordered: $f_1'(x) < \dots < f_n'(x)$.
Some of the first polynomials are:
- $f_1(x) = x^2$, $f_1'(x) = 2x$, $x_1 = 0$
- $f_2(x) = 2x^3 - x^2$, $f_2'(x) = 6 x^2 -2x$, $x_2 = \frac{1}{2}$
- $f_3(x) = 6 x^4 - 6x^3 + x^2$, $f_3'(x) = 24x^3-18x^2+ 2x$, $x_3 \approx 0.7887$
- $f_4(x) = 24 x^5 - 36 x^4 + 14 x^3 -x^2$, $f_4'(x) =120x^4-144x^3 +42x^2 -2x$, $x_4 \approx 0.9082$
Therefore $f_2'(x)-f_1'(x) = 6x^2-4x \geq 0$ for all $x \geq \frac{2}{3}$. Note that, $x_2 < \frac{2}{3} < x_3$.
For $f_3'(x) - f_2'(x) = 24 x^3-24 x^2 + 4 x \geq 0$ for all $x \geq 0.7887$. Here it turns out that at $x_3$ the inequality holds as an equality (coincidence perhaps?), but of course then for $x_4 > x_3$ it holds as a strict inequality.