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 $a>0$?
 A: Since the square root is increasing, it is sufficient to show that $2t \log \log (\frac{1}{t})$ is increasing. Rewrite this as  $\log ((\log (\frac{1}{t}))^{2t})$. Since log is increasing, it is sufficient to show that $(\log (\frac{1}{t}))^{2t}$ is increasing. This has a nicer derivative, namely:
$$2 \left[  \log \left(\frac{1}{t}\right) \log \left(\log \left(\frac{1}{t}\right)\right) -1 \right]\left(\log \left(\frac{1}{t}\right)\right)^{2t-1}$$
A: Taking the derivative of $f(t)=\sqrt{2t\ln{\ln{\dfrac{1}{t}}}}$ with respect to $t$:
$$f'(t)=\dfrac{1}{2\sqrt{2t\ln{\ln{\dfrac{1}{t}}}}}\cdot\left(2\ln{\ln{\dfrac{1}{t}}}+2t\cdot\dfrac{1}{\ln{\dfrac{1}{t}}}\cdot t\cdot\dfrac{-1}{t^2}\right)$$
$$\therefore f'(t)=\dfrac{\ln{\ln{\frac{1}{t}}}-\dfrac{1}{\ln{\frac{1}{t}}}}{\sqrt{2t\ln{\ln{\frac{1}{t}}}}}$$
$f(t)$ is increasing when $f'(t)>0$:
$$\large 0<t< e^{-e^{\mathcal{W}(1)}}: f'(t)>0$$
where $\mathcal{W}$ is the Lambert W function, which gives (numerically):
$$\large e^{-e^{\mathcal{W}(1)}}\simeq 0.17149128425$$

Working out $f'(t)=0$:
$$\ln{\ln{\frac{1}{t}}}-\dfrac{1}{\ln{\frac{1}{t}}}=0\Longleftrightarrow\ln{\dfrac{1}{\ln{\frac{1}{t}}}}+\dfrac{1}{\ln{\frac{1}{t}}}=0\Longleftrightarrow\dfrac{1}{\ln{\frac{1}{t}}}\cdot e^{\frac{1}{\ln{\frac{1}{t}}}}=1\Longleftrightarrow\dfrac{1}{\ln{\frac{1}{t}}}=\mathcal{W}(1)$$$$\large\therefore t=e^{-\frac{1}{\mathcal{W}(1)}}=e^{-e^{\mathcal{W}(1)}}\simeq 0.17149128425$$
A: 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 $f(x)=\frac 1x -\log(x)$. 
A: 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 $\,\ln u >1$.
A: I take log (not ln) as the question does. Function log log $\frac{1}{x}$ has the open interval (0, 1) as domaine and is decreasing; it is positive over (0, $\frac{1}{10}$) and negative otherwise. It follows the domaine of the given function is (0, $\frac{1}{10}$). Since the continuous (on its domaine) given function tends to zero when x tends to zero (since $\frac{1}{x}$ is of greater order than log log $\frac{1}{x}$) and f($\frac{1}{10}$) = 0 and because the square root is increasing, the given function is increasing from 0 to the maximum of f wich exists for a $x_0$ with 0 < $x_0$ < $\frac{1}{10}$. 
The asked a can be all element of the open interval (0, $x_0$).
