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?
 A: Hint. If you switch variable to $y=\ln x$, you want to prove that
$$ \ln(y)^2 < y $$
for sufficiently large $y$. Does that seem easier to prove?
If not, then switch variables once again to $z = \ln y$ and prove
$$ z^2 < e^z $$
for sufficiently large $z$.
A: 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 $x>\mathrm e^4$.
A: 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.
A: $y:=\log (\log x)$, $x$ large ;
Take logs:
LHS:
$2\log (\log (\log x))=2 \log y$
RHS:
$\log (\log x)=y$
Need to show :
$2 \log y < y$ for sufficiently large $y$.
Used: $\log$ is an increasing function.
A: For $x > 1$,
and for $0 < a < 1$,
$\begin{array}\\
\ln(x)
&=\int_1^x \dfrac{dt}{t}\\
&<\int_1^x \dfrac{dt}{t^{1-a}}\\
&=\int_1^x t^{a-1}dt\\
&=\dfrac{t^a}{a}|_1^x\\
&=\dfrac{x^a-1}{a}\\
&<\dfrac{x^a}{a}\\
\end{array}
$
If $a = \frac12$,
this gives
$\ln(x)
\lt 2x^{1/2}
$.
If $a = \frac14$,
this gives
$\ln(x)
\lt 4x^{1/4}
$
so
$\dfrac{\ln(x)}{x^{1/2}}
\lt 4x^{-1/4}
$.
Therefore
$\dfrac{\ln(\ln(x))}{\ln(x)^{1/2}}
\lt \dfrac{4}{\ln(x)^{1/4}}
$.
A: For real $t > 0,$ we have
$$  t^{1/e} \geq \ln t,   $$
with equality only at $t = e^e.$ That is nice and specific, you should confirm that, any way you can. Draw some graphs, at least. Note that $e^e \approx 15.154$
Next, when $t > 1,$ we have $\sqrt t > t^{1/e}.$ 
Finally, take $t = \ln x$ for $x > 1$
The inequality you want is ALways true. not just for big numbers
A: You should always remember the fundamental inequality satisfied by log function $$\log x\leq x-1,\,x>0\tag{1}$$ with equality happening only when $x=1$. Let $x>1$ and then replacing $x$ by $\sqrt[4]{x} $ we can see that $(1)$ implies $$\log x<4\sqrt[4]{x}$$ or $$\log^2x<16\sqrt{x}$$ and clearly $16\sqrt{x}<x$ if $x>256$. Thus we have $$\log^2x<x$$ if $x>256$. Replacing $x$ by $\log x$ we see that $$(\log\log x) ^2<\log x$$ if $x>e^{256}$.
The use of L'Hospital's Rule is fine if it is allowed by your professor (some people bar it for many exercises). On the other hand it is much simpler to prove the limit $$\lim_{x\to\infty} \frac{(\log x) ^a}{x^b}=0,\,a>0<b\tag{2}$$ using inequality $(1)$ and squeeze theorem. 
