<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] Using my answer to your previous question, we have (after trigonometric simplifications)
$$a=\frac{4}{3} \cos ^2\left(\frac{1}{6} \left(\cos
   ^{-1}\left(\frac{27 s}{2}-1\right)+2 \pi \right)\right)$$
$$b=\frac{4}{3} \sin ^2\left(\frac{1}{6} \left(\cos
   ^{-1}\left(\frac{27 s}{2}-1\right)+\pi \right)\right)$$
Then
$$(1-a)^2+2a^2=\frac{1}{3} \left(5-4 \cos \left(\frac{1}{3} \cos
   ^{-1}\left(1-\frac{27 s}{2}\right)\right)+2 \cos
   \left(\frac{2}{3} \cos ^{-1}\left(1-\frac{27
   s}{2}\right)\right)\right)$$
and
$$(1-b)^2+2b^2=\frac{1}{3} \left(5+4 \sin \left(\frac{1}{3} \sin
   ^{-1}\left(1-\frac{27 s}{2}\right)\right)-2 \cos
   \left(\frac{2}{3} \sin ^{-1}\left(1-\frac{27
   s}{2}\right)\right)\right)$$ So, if
$$\Delta=\Big[(1-b)^2+2b^2\Big]-\Big[(1-a)^2+2a^2\Big]$$
$$\Delta=\frac{8}{\sqrt{3}}\sin ^2\left(\frac{1}{6} \cos ^{-1}\left(\frac{27
   s}{2}-1\right)\right) \sin \left(\frac{1}{3} \cos
   ^{-1}\left(\frac{27 s}{2}-1\right)\right) $$
$\Delta$ has the same sign as $\sin \left(\frac{1}{3} \cos
   ^{-1}\left(\frac{27 s}{2}-1\right)\right) $ which varies between $\frac{\sqrt{3}}{2}$ and $0$.
So $$\color{red}{\Delta ~\geq~  0 \quad \forall s \in \left(0,\frac 4{27}\right)}$$
</s>