Maximize $x^{2(1-x)^2}+(1-x)^{2x^2}$ on $[0,1]$ It's a problem of mine :

Let $0<x<1$ and defines the function :
$$f(x)=x^{2(1-x)^2}+(1-x)^{2x^2}$$
We call this function the Napoleon's hat. Now we want to know the maximal height of Napoleon's hat on $[0,1]$. I suspect that the value is $\sqrt{2}$ at $x=0.5$ .

The expression of the derivative is :
$$f'(x)= x^{2 (1 - x)^2} \left(\frac{2 (1 - x)^2}{x} - 4 (1 - x) \log(x)\right) + (1 - x)^{2 x^2} \left(4 x \log(1 - x) - \frac{2 x^2}{(1 - x)}\right)$$
My question :
Can someone show that the maximal value is $\sqrt{2}$ at $x=0.5$ ?
 A: Define $g(x)=x^{2(1-x)^2}$ and $h(x)=-(1-x)^{2x^2}$, so that $f(x)=g(x)-h(x)$. As Daniyar Aubekerov commented, it suffices to show that $f'(x)>0$ for $x\in(0,\frac12)$, which is equivalent to showing that $g'(x)>h'(x)$ for $x\in(0,\frac12)$. We have
\begin{align*}
g'(x) &= 2 (1-x) x^{2(1-x)^2-1} (1-x-2 x \log x) \\
h'(x) &= 2 x (1-x)^{2 x^2-1} (x-2 (1-x) \log (1-x),
\end{align*}
each factor of which is positive; therefore it suffices to show that their quotient $g'(x)/h'(x)$ is greater than $1$. Define
\begin{align*}
a(x) &= (1-x^2) \log (1-x)+(x-2) x \log x \\
b(x) &= \frac{1-x-2 x \log x}{x-2 (1-x) \log (1-x)},
\end{align*}
so that $g'(x)/h'(x) = e^{2a(x)}b(x)$; it suffices to show that $a(x) > 0$ and $b(x) > 1$ for $x\in(0,\frac12)$.
Showing $a(x)>0$: It is easy to check that $a'''(x) > 0$ for $x\in(0,\frac12]$ and that $a''(\frac12)=0$. Therefore $a''(x)<0$ for $x\in(0,\frac12)$, and hence $a(x)$ is concave on $(0,\frac12)$. Since $a(0)=a(\frac12)=0$, it follows that $a(x)>0$ for $x\in(0,\frac12)$.
Showing $b(x)>1$: If we further define $c(x) = 1-x-2 x \log x$ and $d(x) = x-2 (1-x) \log (1-x)$, so that $b(x) = c(x)/d(x)$, it suffices to show that $c(x) - d(x) > 0$ for $x\in(0,\frac12)$. But it is easy to check that $c''(x)-d''(x)<0$ for $x\in(0,\frac12)$ and hence that $c(x)-d(x)$ is concave on $(0,\frac12)$; since $c(0)-d(0)=1$ and $c(\frac12)-d(\frac12)=0$, the difference $c(x)-d(x)$ is positive for $x\in(0,\frac12$).
