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?

In the $\varepsilon$-$\delta$ definition of $$\lim_{x\rightarrow \infty} \frac{(\ln(\ln(x)))^2}{\ln(x)}$$
take $\varepsilon = 1$ and have some $M \in \mathbb{R}$ such that $$\forall x \geq M \ \ \left| \frac{(\ln(\ln(x)))^2}{\ln(x)} \right| < 1.$$
For large $x \in \mathbb{R}$ there is no need for the absolute value, so we have $$\forall x \geq M \ \ \frac{(\ln(\ln(x)))^2}{\ln(x)} < 1,$$ which gives the result.
