Let be the function :
$$g(x,y)=\ln\left(\left|\frac{\left(2-\left(1-\frac{1}{x}\frac{2y}{1-y}\right)^{x}\right)}{2}+\frac{\left(1+\frac{\frac{1}{x}2y}{1-y}\right)^{x}}{2}\right|\right)+\ln\left(\left|\frac{\left(2-\left(1-\frac{2yx}{1-y}\right)^{\frac{1}{x}}\right)}{2}+\frac{\left(1+\frac{x2y}{1-y}\right)^{\frac{1}{x}}}{2}\right|\right)$$
Conjecture
Then $a\in(-0.2,0.2),\exists \varepsilon,0<\varepsilon<1/2$:
$$g\left(x,a\right)+g\left(x,-a\right)\geq 0, x\in[1-\varepsilon,1+\varepsilon]$$
I try to show that the function is convex then use the tangent line method as $f(1)=f'(1)=0$ but the second derivative is horrendous .
As other remark the case $x=1$ is similar to the application of Bernoulli's inequality on the exponents .
As other attempt we have :
$$\left|h\left(x,a\right)-h\left(1,a\right)\right|\leq \left|g\left(x,a\right)-g\left(1,a\right)\right|$$
Where :
$$h(x,y)=\ln\left(\left|\frac{\left(2-\left(1-\frac{1}{x}\frac{2y}{1-y}\right)^{x}\right)}{2}+\frac{\left(1+\frac{\frac{1}{x}2y}{1-y}\right)^{x}}{2}\right|\right)\ln\left(\left|\frac{\left(2-\left(1-\frac{2yx}{1-y}\right)^{\frac{1}{x}}\right)}{2}+\frac{\left(1+\frac{x2y}{1-y}\right)^{\frac{1}{x}}}{2}\right|\right)$$
Question :
How to (dis)prove it ?
My strategy :
As we have with the constraint above :
$$|g(x,a)|<1$$
Then with the kind of inequalities $x,y\in[-1,1]$:
$$\left|xy-1\right|<\left|x+y-2\right|$$
We establish the inequality $\operatorname{I}$ :
So we have :
$$\left|h\left(x,a\right)-h\left(1,a\right)\right|\leq \left|g\left(x,a\right)-g\left(1,a\right)\right|$$
Or :
$$h\left(x,a\right)-h\left(1,a\right)+h\left(x,-a\right)-h\left(1,-a\right)\leq \left|g\left(x,a\right)-g\left(1,a\right)\right|+\left|g\left(x,-a\right)-g\left(1,-a\right)\right|$$
Then it seems we have :
$$r\left(x\right)=\left|g\left(x,a\right)-g\left(1,a\right)\right|+\left|g\left(x,-a\right)-g\left(1,-a\right)\right|<\frac{1}{|a|}\left(g\left(x,a\right)+g\left(x,-a\right)\right)$$