For which $\alpha$ is $x\le 1/|\log(x)|^\alpha$ with $x\in(0,1)$? Let $x\in(0,1)$. I want to find for which $\alpha>0$ it's true that
$$
x\le\frac{1}{|\log(x)|^\alpha}
$$
Plotting the functions $x\mapsto x$ and $x\mapsto 1/|\log(x)|^\alpha$ on WolframAlpha, I can see the inequality holds, at least for $\alpha$ approximately less than 2 or so. But it must be possible to prove something analytically.
So far, using that $\log(x)<0$,  I have been able to derive:
\begin{align*}
&x\le-(\log(x))^{-\alpha}\iff x^{-1}\ge -(\log(x))^\alpha\iff x^{-\frac{1}{\alpha}}\ge-\log(x)
\\
\iff&-\frac{1}{\alpha}\log(x)\ge\log(-\log(x))
\\
\iff&-\frac{1}{\alpha}\ge\frac{\log(-\log(x))}{\log(x)}
\\
\iff&-\alpha\le\frac{\log(x)}{\log(-\log(x))}
\\
\iff&\alpha\ge-\frac{\log(x)}{\log(-\log(x))}.
\end{align*}
Am I missing something here? After all, based on the numerical plots I have seen, I should have had a bound for $\alpha$ from above; not below.
 A: For $x∈ (0,1)$
$$
(1)\qquad x ≤ \frac{1}{\left|\ln x\right|^\alpha} \iff x\,(-\ln x)^\alpha ≤ 1.
$$
Let $f(x) = x\,(-\ln x)^\alpha$. Since $f\underset{0}{\rightarrow} 0$ and $f(1) = 0$, $f>0$ on $(0,1)$ and $f$ is $C^1$, we know its maximum is reached on some $x_0\in(0,1)$ such that $f'(x_0)=0$. This implies
$$
(-\ln x_0)^\alpha - \alpha\,(-\ln x_0)^{α-1} = 0.
$$
so that $-\ln x_0 = \alpha$. Therefore, $x_0 = e^{-\alpha}$ and the maximum is
$$
f(x_0) = x_0\,(-\ln x_0)^\alpha = e^{-\alpha}\,\alpha^\alpha = (\tfrac{\alpha}{e})^\alpha.
$$
We deduce that $(1)$ is true for every $x∈ (0,1)$ if and only if $(\tfrac{\alpha}{e})^\alpha \leq 1$ or equivalently
$$
α≤e.
$$
A: Since $\log x < 0$ in this interval, the inequality becomes
$$
x \leq (-\log x)^{-\alpha} \Leftrightarrow \log x \leq -\alpha \log (-\log x) 
$$
and, since the sign of $\log (-\log x)$ changes at $x=1/e$, we must have
$$
\begin{cases}
-\alpha \ge \frac{\log x}{\log(-\log x)}, & x < 1/e\\
-\alpha \leq \frac{\log x}{\log(-\log x)}, & x > 1/e
\end{cases}
$$
In order for the inequality to hold for all $x < 1/e$ we must have that
$$
\alpha \leq \frac{\log x}{\log(-\log x)}, \quad \forall x \in (0,1/e).
$$
Since the function attains a minimum value of $e$, we must have $\alpha \leq e$.
Similarly, in order for the inequality to hold for $x< 1/e$, we must have $\alpha \ge 0$.
