Function grows slower than $\ln(x)$

For $x>1$, $x \mapsto \ln(\ln(x))$.

$$f(x)=\sqrt{\ln x}$$

should do, for $x\ge 1$.

If we want to really go to the extreme, we can look at some smooth continuation of the Inverse Ackermann function.

Also, we can look at the inverse of a transexponential function. A function, $f(x)$ is transexponential if there exists some there exists points $p_1, p_2,..., p_n,...$ such that $\forall x > p_1$ $f(x) > e^x$, $\forall x > p_2$, $f(x) < e^{e^x}$, $\forall x > p_3$ $f(x) > e^{e^{e^x}}...$ The inverse of this $f(x)$ will grow very slowly.