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 $a>0$?

First the function is defined if 


*

*$t>0$ (because of $\,\ln\dfrac1t$);

*$\ln\smash[t]{\dfrac1t}>0$ (because of the next $\,\ln$);

*$\ln\dfrac1t>1$ (because of the radicand and $t>0$).
Finally all this amounts to $\dfrac1t>\mathrm e\iff 0 < t <\dfrac1{\mathrm e}$


Second, showing the function is increasing amounts to showing $t\ln\ln\dfrac1t$ is, or, if we set $u=\dfrac1t$, to showing  $f(u)=\dfrac{\ln\ln u}{u}$ is a decreasing function of $u$. This means that  for $u>\mathrm e$, the logarithmic derivative of the numerator is ${}<{}$ logarithmic derivative of the denominator, i. e.:
$$\frac1{u\ln u}<\frac1u$$
which is true because $\,\ln u >1$.
