Given a small number $\varepsilon >0$ and a constant $1/3\le \alpha < 1 $, I am looking for the smallest possible number $x^*$ such that for all real $x\ge \max\{x^*,3\}$, we have $$\frac{x}{(\log x)^{\alpha}} \ge \varepsilon^{-2\alpha}. \tag1$$
Equivalently, the quantity I am looking for is (an upper bound of) the solution of the above equation with the $``\ge" $ symbol replaced by $``="$, but my understanding is that the equation is not tractable (WolframAlpha is also clueless), and I am not sure of how to find bounds for the solution. For instance, when I look for candidates of the form $x^*\equiv\varepsilon^\beta$, I end up having to solve $(\log\varepsilon)^{-\alpha}\ge\beta^\alpha\varepsilon^{-2\alpha-\beta}$ for $\beta$, which does not seem easier.
Since for the values of $x$ and $\alpha$ I'm considering, we have $(\log x)^{-\alpha}\ge(\log x)^{-1}$, I decided instead to let $M\equiv\varepsilon^{-2\alpha} $ and look for the solution of the easier equation $$ \frac{x}{\log x} \ge M\tag2$$ For which WolframAlpha tells me the solution is given by $$x^* = \exp(-W(-1/M)) = \exp(-W(-\varepsilon^{2\alpha})) \tag3 $$ where $W$ is Lambert's $W$ function (not sure how that result is derived btw).
To finish, I just need a good upper bound on $x^*$ above but 1) I am not really familiar with the $W$ function and, searching online, I can't find good bounds in the interval $[-1/e,0[ $, and 2) the solution $(3)$ is suspicious : as $\varepsilon\to 0$, the solution $x^*$ should diverge, but according to $(3)$, $x^*$ converges to $0$ ! I am not sure of what went wrong, and the mistake probably comes from my lack of understanding of the $W$ function.
Hence, I ask can anyone give an estimate of the smallest solutions of $(1)$ and/or $(2)$ in terms of $\boldsymbol{\varepsilon}$ and $\boldsymbol{\alpha}$ ? I'd prefer avoiding the use of Lambert's function, if possible.
Thank you for your time !
Let $y := x^{1/\alpha}$. The condition is written as: for all $y^\alpha \ge \max(x^\ast, 3)$, $$\frac{y^\alpha}{(\alpha \ln y)^\alpha} \ge \varepsilon^{-2\alpha}$$ or $$\frac{y}{\ln y} \ge \frac{\alpha}{\varepsilon^2}. \tag{1}$$
Note that $\frac{y}{\ln y} \ge \frac{3^{1/\alpha}}{\ln 3^{1/\alpha}}$ on $y \ge 3^{1/\alpha}$. If $\frac{\alpha}{\varepsilon^2}\le \frac{3^{1/\alpha}}{\ln 3^{1/\alpha}}$, then (1) is true for all $y\ge 3^{1/\alpha}$ (thus, $x^\ast = -\infty$). In the following, assume that $\frac{\alpha}{\varepsilon^2} > \frac{3^{1/\alpha}}{\ln 3^{1/\alpha}}$ (in this case, $x^\ast > 3$). The condition (1) is equivalent to $$y \ge - \frac{\alpha}{\varepsilon^2} W_{-1}\left(-\frac{\varepsilon^2}{\alpha}\right)$$ where $W_{-1}(\cdot)$ is the second branch of the Lambert W function. Thus, $$x^\ast = \left(- \frac{\alpha}{\varepsilon^2} W_{-1}\left(-\frac{\varepsilon^2}{\alpha}\right)\right)^{\alpha}.$$