How to show this crazy inequality of logarithms and constant number? Is there any way to solve this inequality? I asked my friend for help, but he couldn't do it. I can't use even derivatives and his solution was including them. So, after many transformations i have to show this inequality : 
$$\frac{\ln x}{t} + \ln t > \ln \ln \ x$$ for $t > 1$
Is there a way to show it's true ? ( Wolfram says so... ). But if i over-complicated things, here's the original inequality : $$t \ln x < x^{1/t}$$
I have to show it's true for some large $x$. At my first inequality, it must be true for $x > 0$. If anyone would be interested in those weird transformations, i'll provide them.
I've spent over 3 hours in one go thinking about it, but i always fail... 
Would be great to receive some cool hints or answer to this task. 

Find the minimum in $t$ of $\frac{\log(x)}{t}+\log(t)$. Its derivative is $0$ when $t=\log(x)$. Thus,
$$
\begin{align}
\frac{\log(x)}{t}+\log(t)
&\ge\frac{\log(x)}{\log(x)}+\log(\log(x))\\
&=1+\log(\log(x))\\[4pt]
&\gt\log(\log(x))
\end{align}
$$

Without derivatives:
We have the inequality for all $y$: $1+y\le e^y$. If we let $y=-\log(u)$, we get
$$
1-\log(u)\le\frac1u
$$
Consider $t=u\log(x)$. Then we have
$$
\begin{align}
\frac{\log(x)}{t}+\log(t)
&=\frac1u+\log(u)+\log(\log(x))\\
&\ge1+\log(\log(x))
\end{align}
$$
