Is $\sqrt{2t \log{\log{\frac{1}{t}}}}$ increasing for $t\in(0,a)$, for a suitable $a>0$? I believe it's obviously, but I tried a lot, and have no clue how can I show that
$$\sqrt{2t \log{\log{\frac{1}{t}}}}$$
is increasing for $t\in(0,a)$ for a suitable $a>0$?

Hint
Quoting wythagoras, "Since the square root is increasing, it is sufficient to show that $t \log \log (\frac{1}{t})$  is increasing".
Now, change variable $t=e^{-x}$ and work now with $e^{-x}\log(x)$ for which the beautiful derivative is $$\frac{e^{-x} (1-x \log (x))}{x}$$ In such a way, you are only concerned by the sign of $f(x)=\frac 1x -\log(x)$. 
