Solving for $x$ such that $x^2 \log(\alpha/x) \leq \beta$ where $\alpha, \beta >0$. I am trying to solve the following inequality for $x$. Essentially it suffices even if I can find one value of $x$ which satisfies this:
$$x^2 \log\left(\frac{\alpha}{x}\right) \leq \beta,$$
where $\alpha, \beta >0$. How can I approach it?
Some background on where the problem is coming from:
Basically, I have the following inequality:
$$\beta\left(cH\sqrt{d\log\left(\frac{Hkd}{\beta\delta\delta_1}\right)}+\frac{4}{3}\right) \leq \frac{1}{4}$$
and I need to find a positive value for $\beta$ for which the inequality satisfies. Here $c,H,d, k,\delta, \delta_1$ are positive constants.
 A: Given the context, the correct inequality to consider is
$$x\sqrt{\log \frac Ax} \le B$$
which is slightly different, only because we need to restrict $\log \frac Ax\ge 0$ i.e. $x\le A$.
Without loss of generality, we may assume $A=1$ by replacing $X = x/A$, to find the inequality
$$ X \sqrt{\log \frac 1X} \le \frac BA$$
Then if we know that $X\le C$ implies $X \sqrt{\log \frac 1X}\le B/A$, $x\le AC$ implies $ x\sqrt{\log \frac Ax} \le B$.
So from now on assume $A=1$ and $x\le 1$. On this region, $x^2\log \frac 1x$ has a maximum of $\frac1{2e}$ at $x=e^{-1/2}$. So we can take any $0\le x\le 1$ if $B\ge \frac1{\sqrt{2e}}$. Here's a plot of $x^2 \log(\frac Ax)$. (Note, $A\neq 1$ in the graph)

If $B\le 1/\sqrt{2e}$ then we can instead try to look for a positive solution near $x=0$, since the graph shows us that $x^2\log \frac 1x$ is small when $x\approx 0$. Let's say $\log$ means the natural log.
Observe that $-\log x\le 1/x$, because $\log x \ge -1/x$ because $x\ge e^{-1/x}$ because $1/t \ge e^{-t}$ ($t\ge 1$) because $e^t \ge t$. This means that
$$ x^2\log \frac 1x = -x^2\log x  \le  x .$$
Hence, if we want $x>0$ such that $x^2 \log \frac1x \le B^2$, it is enough to take  $ x \in (0,  B^2)$.

Purple is $y=B^2$; the green line $y=x$ gives us a sufficient condition (shaded red) where $x\sqrt{\log \frac 1x} \le B$. Interactive Desmos graph link here.
In the general case $A\neq 1$, it follows that if $0< x < \min(A,B^2/A)$, then $x\sqrt{\log \frac Ax} < B$.
A: Since you are looking for just one solution, here's one.
We have $\alpha, \beta>0$. Now look at any $x\ge\alpha$. Then $\ln(\alpha/x)\le0\Rightarrow x^2\ln(x/\alpha)\le 0$ and thus $$x^2\ln(x/\alpha)\le 0<\beta$$ and thus for $\beta>0$, any $x\ge \alpha$ is always a solution.
