How can you maximize $f(x) = W(x) - (\ln(x) - \ln{\ln{x}})$ for $x\geq 2$?
Numerically the answer seems to be at around $x \approx 41$ where you get $f(41) \approx 0.31$. Mathematica suggests the maximum is at $x= e^{1+e}$.
$W$ is the Lambert-W function.
Using the fact that
$$ W'(x) = \frac{W(x)}{x(1+W(x))} $$
we calculate
$$ f'(x) = \frac{W(x)+1-\ln x}{x\ln x(1+W(x))}, $$
so to optimize $f$ we want to solve the equation
$$ W(x)+1 = \ln x. $$
Exponentiating both sides yields
$$ e e^{W(x)} = x, $$
and after multiplying both sides by $W(x)$ this becomes
$$ e W(x) e^{W(x)} = x W(x). $$
Since $W(x)e^{W(x)} = x$ this is equivalent to
$$ ex = xW(x) $$
or just
$$ W(x) = e. $$
Thus
$$ x = ee^e = e^{1+e}. $$
Plugging this back into the function we get
$$ f(e^{1+e}) = \ln(1+e) - 1. $$