Let $a=1,b=1.1,p=\frac{4c}{\left(a-b\right)}\sqrt{\frac{1-\ln2}{1+\ln2}},c=d$ then $\exists d \in(0,0.1)$ such that for $x>0$:
$$\frac{ax^{p}+b}{cx^{p}+d}-\frac{a+b}{c+d}+2\geq \left(1+x\right)^{x^{-\frac{1}{2\ln2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)}}\tag{I}$$
Simply $d$ is choose as $f'''(1)=g'''(1)$ where :
$$\frac{ax^{p}+b}{cx^{p}+d}-\frac{a+b}{c+d}+2=f(x),g(x)=\left(1+x\right)^{x^{-\frac{1}{2\ln2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)}}$$
For a motivation see this link Find the "best" $n$ such that $g(x)\leq 4$
I spend more than a day to find a proof of this fact .
Logarithmic derivative doesn't helps here and I have tried a lot of things including Padé approximation .
Question :
How to show it ?
Any clue is very very welcome .
Ps: I cannot find $d$ can someone can give the expected value in a comment thanks !
$$g'''(1)=a=\left(2\left(\frac{1}{2}+\frac{1}{2}\left(\sqrt{\frac{1-\ln2}{1+\ln2}}-1\right)\right)^{3}+6\left(\frac{1}{2}+\frac{1}{2}\left(\sqrt{\frac{1-\ln2}{1+\ln2}}-1\right)\right)\left(-\frac{1}{4}-\frac{\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)}{2\ln2}-\frac{1}{2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)\left(-1-\frac{\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)}{2\ln\left(2\right)}\right)\right)+2\left(\frac{1}{4}+\frac{3\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)}{8\ln\left(2\right)}-\frac{3\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)}{4\ln2}\cdot\left(-1-\frac{\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)}{2\ln2}\right)-\frac{1}{2}\left(1-\sqrt{\frac{1-\ln2}{1+\ln2}}\right)\left(-2-\left(\frac{1-\sqrt{\frac{1-\ln2}{1+\ln2}}}{2\ln2}\right)\right)\cdot\left(-1-\left(\frac{1-\sqrt{\frac{1-\ln2}{1+\ln2}}}{2\ln2}\right)\right)\right)\right)$$
That's why I say it's a bit hard to show .
