How to prove that $\left(\ln(\ln(x)) \right)^2 \lt \ln(x)$ How to prove that $\left(\ln(\ln(x)) \right)^2 \lt \ln(x)$ for sufficiently large $x$
This is what I did. Using L'Hopital's rule we have 
 $$\lim_{x\to\infty}\frac{\left(\ln(\ln(x)) \right)^2 }{ \ln(x)}=0$$
So this implies that $\left(\ln(\ln(x)) \right)^2 \lt \ln(x)$
Is that enough?

Hint:
Star from an inequality traditionally used in high-school to prove that $\;\lim_{x\to+\infty}\dfrac{\ln x}x=0$:
$$\ln x<\sqrt x\quad\forall x>4,$$
and replace $x$ with $\ln x$: if $\ln x>4$, then
$$\ln(\ln x)<\sqrt{\mkern1mu\ln x\mathstrut}$$
Note that, as $x>4>\mathrm e$, both sides of the inequality are positive, hence you can square to obtain the required inequality for all $x>\mathrm e^4$.
