Problem inspired by Vasile Cirtoaje :
Let $0<b<0.5<a<1$ such that $a+b=1$ then we have :
$$\Re\Bigg(\Bigg(1-\frac{W(-\ln(2a))}{\ln(2a)}\Bigg)^{1-\frac{1}{a}}+\Bigg(1-\frac{W(-\ln(2b))}{\ln(2b)}\Bigg)^{1-\frac{1}{b}}\Bigg)<1$$
Where we speak about the Lambert's function and the real part of a complex number .
Well first of all we have in other word :
$$(2a)^{(2a)^{(2a)\cdots}}=\frac{W(-\ln(2a))}{-\ln(2a)}$$
I have tried to prove something like :
$\exists x \quad $$0<x<0.5$ such that : $$\Re\Bigg(\Bigg(1-\frac{W(-\ln(2x))}{\ln(2x)}\Bigg)^{1-\frac{1}{x}}\Bigg)<x$$
And
$\exists x \quad $$0.5<x<1$ such that : $$\Re\Bigg(\Bigg(1-\frac{W(-\ln(2x))}{\ln(2x)}\Bigg)^{1-\frac{1}{x}}\Bigg)<x$$
I have tried also an approximation with some parabola or :
$\exists x \quad $$0.5<x<1$ such that : $$\Re\Bigg(\Bigg(1-\frac{W(-\ln(2x))}{\ln(2x)}\Bigg)^{1-\frac{1}{x}}\Bigg)<ax^2+bx+c$$
Without success because due to the divergence I don't know how to deal with the real part .
A relevant problem is If $0<a<1, 0<b<1$, $a+b=1$, then prove that $a^{2b}+ b^{2a} \le 1$
Well if you have a way or something plausible feel free to make comments/answers .
Thanks in advance !