Is this quadratic polynomial monotone at solutions of this cubic?

Question

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.

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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.

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The motivation for this problem come from this question.

Answer 1

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.

Answer 2

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 $$\color{red}{\Delta ~\geq~ 0 \quad \forall s \in \left(0,\frac 4{27}\right)}$$