<s>[INST] Is this quadratic polynomial monotone at solutions of this cubic? Let $0<s < \frac{4}{27}$. The equation $x(1-x)^{2}=s$ admits exactly two solutions in $(0,1)$: Denote by $a,b$ be these solutions, and suppose that $a<b$.
Does
$$
(1-a)^2+2a^2<(1-b)^2+2b^2
$$
hold?
The limitation on the range of $s$ is due to $$\max_{a \in (0,1)}a(1-a)^{2}=\frac{4}{27},$$
which is attained at $a=\frac{1}{3}$.
I checked this numerically. I hope for an analytic argument which does not require solving explicitly the cubic.

Here is some partial information:
Define $F(x)= (1-x)^2+2x^2$. Then
$$
F'(x)=2(3x-1)>0 \iff x > \frac{1}{3}.
$$
Thus $F|_{(0,\frac{1}{3})}$ is decreasing, while $F|_{(\frac{1}{3},1)}$ is increasing.
Unfortunately, this does not seem to help.

The motivation for this problem come from this question.
 [/INST] The given inequality
$$(1-a)^2 + 2a^2 < (1-b)^2 + 2b^2$$
is equivalent to
$$0 < (b-a)\cdot(3(b+a) - 2)$$
i.e.
$$ b+a > \frac{2}{3}.$$
If $c$ is the third root of the equation $x(1-x)^2 = s$, then Vieta's formula says $a+b+c = 2$, so the inequality is equivalent to
$$ c < \frac{4}{3}.$$
But if $c \ge \frac{4}{3}$, then
$$c(c-1)^2 \ge \frac{4}{3}\left(\frac{4}{3}-1\right)^2 = \frac{4}{27} > s,$$
so no value of $c\ge \frac{4}{3}$ is a root of $x(x-1)^2 = s$.
Therefore the given inequality is true.
</s>