Let $0<x<0.5$ or $0.5<x<1$ defines the following function : $$f(x)=x^{4(1-x)^2}+(1-x)^{4x^2}$$
Claim : The function $g(x)=\exp(|1-f(x)|)-1$ is a logarithmically concave function
Or : The function $h(x)=\ln(g(x))=\ln(\exp(|1-f(x)|)-1)$ is a concave function
The second derivative is very ugly and I prefer this form
Furthermore I try to use the definition of a logarithmically concave function but it gives a lot of difficulties .
I try also the Popoviciu's inequality but it requires to be continuous so I think it collapses .
If someone have a good idea it will be great .
Thanks a lot for your time and patience .
Edit
An example : If we use Popoviciu's inequality where $x=0.5-\varepsilon$ and $y=a$ and $z=0.5-a$ where $a\in[0.1,0.4]$ and $0<\varepsilon<10^{-2}$ we have :
$$\frac{h(0.5-\varepsilon)+h(a)+h(0.5-a)}{3}+h\Big(\frac{0.5-\varepsilon+a+0.5-a}{3}\Big)$$ Or : $$\frac{h(0.5-\varepsilon)+h(a)+h(0.5-a)}{3}+h\Big(\frac{1-\varepsilon}{3}\Big)$$
And :
$$\frac{2}{3}\Big(h\Big(\frac{0.5-\varepsilon+a}{2}\Big)+h\Big(\frac{0.5-\varepsilon+0.5-a}{2}\Big)+h\Big(\frac{0.5-a+a}{2}\Big)\Big)$$
Or:$$\frac{2}{3}\Big(h\Big(\frac{0.5-\varepsilon+a}{2}\Big)+h\Big(\frac{1-\varepsilon-a}{2}\Big)+h\Big(\frac{0.5}{2}\Big)\Big)$$
Now when $\varepsilon$ tends to $0$ the first quantity tends to $-\infty$ and the second is finite so it prove the partially the claim.
I have this :
Defines :$$p(x)=2\sqrt{x^{4(1-x)^2}(1-x)^{4x^2}}$$ And:$$q(x)=\ln(\exp(|1-p(x)|)-1)$$
Then $$h''(x)\leq q''(x)\leq 0$$