How to find $C$ (improving a result due to user Clement.C)?

Question

I improve the result see (Elegant) proof of : $x \log_2\frac{1}{x}+(1-x) \log_2\frac{1}{1-x} \geq 1- (1-\frac{x}{1-x})^2$ :

What is the best constant $C$ such that $x\in(0,0.5]$:

$\frac{1}{\ln\left(2\right)}\left(x\ln\left(\frac{1}{x}\right)+\left(1-x\right)\ln\left(\frac{1}{1-x}\right)\right)+\frac{\ln\left(1-2\left(1-x\right)x\right)}{\ln\left(2\right)}\leq -\left(1-\left(1-\frac{x}{1-x}\right)^{2}+\frac{\ln\left(1-f\left(\frac{1}{2}\right)\left(\frac{1}{2^{C-1}}-x^{C}\right)x^{C}\right)}{\ln\left(2\right)}\right)$

Where :

$$f\left(x\right)=\frac{1}{2}\left(\frac{1}{2^{C-1}}-x^{C}\right)^{-1}x^{-C}$$

It seems there is a problem with the derivatives around zero .So to find $C$ it might be useful to have an asymptotic approximation around this limit of domain .

You can find an attempt of mine in the linked question and other answer which could be useful

Currently $C\simeq 0.76131$.

How to find the $C$ such that we have the inequality ?

Sides notes :

If $C$ as optimal constant does'nt admit a closed form we can attempt to show it for $C=e^{-3/11}$

We can also on the other hand use :

$$x\ln x+(1-x)\ln 1-x \geq\ln \left(1-\sqrt{x(1-x)}\right)$$

Wich can be solved without derivative here New bound for Am-Gm of 2 variables

After some hours it seems we have the equality :

$$\lim_{x\to 0}\frac{\ln\left(C\right)}{f\left(\frac{1}{2}\right)}-\frac{\ln\left(kC\right)}{\ln\left(\frac{k}{d}\right)\cdot f\left(x\right)}-\ln\left(d\right)=\lim_{x\to 0}-\frac{\ln\left(kC\right)}{\ln\left(2\right)\cdot f\left(x\right)}-\frac{\ln\left(kC\right)}{\ln\left(2-\frac{k}{1+d}\right)\cdot f\left(x\right)}=0$$

Where :

$$\ln\left(\frac{k+1}{-\left(1-f\left(\frac{1}{2}\right)\right)}\right)=2,d=0.827...$$

Edit :12/01/2023

Let $d+c=k$ as above then define :

$$h\left(x\right)=\left(\frac{\ln\left(C\right)}{f\left(\frac{1}{2}\right)}-\frac{\ln\left(pC\right)}{\ln\left(\frac{p}{d}\right)\cdot f\left(x\right)}-\ln\left(d\right)\right)$$

Where $p$ is the Catalan's constant plus one or $p=1.915...$

Now define :

$$r\left(x\right)=\sum_{n=1}^{\infty}\frac{1}{e^{x^{2}n}+1+\ln\left(1+e\right)}$$

Then we have :

$$r(c)\simeq h(c)$$

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See also the derivation due to Wolfram Alpha . See also the q-PolygammaFunction

Answer 1

If you minimize $C$ with respect to $(x,C)$ under the equality constraint

$$\color{red}{\large C=0.7613114510607833}$$ which is almost the solution of

$$\log[1+(t-1)^2]=A \qquad \text{with} \qquad t=\left(\frac{6}{125}\right)^C$$ with $$A=\frac{1725473 \log (2)}{1860500}+\frac{\log (27)+122 \log (61)}{125}-\log \left(\frac{14893}{250}\right) $$

The solution of the above equation is $$C=\frac{\log \left(1-\sqrt{e^A-1}\right)}{\log \left(\frac 6{125}\right)}$$ and its numerical value is $$C=0.76131146045\cdots$$

All of this was done using a series expansion around $x=\frac{3}{125}$ which is close to the result of the minimization.

If the expansion is done around $x=\frac{24001}{1000000}$ instead of $x=\frac{24000}{1000000}$, the solution of the corresponding equation would be $$C=0.76131145633\cdots$$

Edit

It is possible to expand around $x=0$ $$G=\text{(lhs - rhs)}\log(2)=\log \left(1-\frac{1}{2} (2 x)^C \left(2-(2 x)^C\right)\right)+$$ $$x\log \left(\frac{4}{e x}\right)+\sum_{n=2}^p a_n\,x^n+O(x^{p+1})$$ The first $a_n$ are $$\left\{\frac{2 \log (2)-1}{2},\frac{7}{6},\frac{23-12 \log (2)}{12} ,\frac{31-40 \log (2)}{20},-\frac{1+90 \log (2)}{30}\right\}$$

Minimizing $G^2$ with respect to $(x,C)$ for a given $p$ to any required accuracy does not present any difficulty.

Ignoring the summation already gives $x=0.0257$ and $C=0.7618$.

Thanks to Wolfram Alpha $$\color{blue}{40\, C \sim 8 \binom{\Omega}{\Omega!}+16 \binom{\Omega!}{\log (\Omega)}+6 \binom{\log (\Omega)}{\Omega!}+}$$ $$\color{blue}{101 \binom{\Omega}{\log (\Omega)}-29 \binom{\Omega!}{\Omega}-58 \binom{\log (\Omega)}{\Omega}}$$ gives all figures.