I was wondering if there is an analytical solution for the following inequality:
$ax^2 \le \log(bx), \quad x,a,b>0$
where $a,b$ are constants. It seems that this problem involves Lambert W-function, but it is still not very clear to me if there is an analytical solution in the form of
$x \le C(a,b),$
where $C(a,b)$ is the upper bound of $x$ determined by $a$ and $b$.
Thanks a lot!
On
Consider the function and its derivatives $$f(x)=a x^2-\log (b x)$$ $$f'(x)=2 a x-\frac{1}{x}$$ $$f''(x)=2 a+\frac{1}{x^2}$$ Since $a$ and $b$ are positive, the first derivative cancels at a point $$x_*=\frac{1}{\sqrt{2a} }\implies f(x_*)=\frac{1}{2}-\log \left(\frac{b}{\sqrt{2a} }\right)$$ and the second deivative is stricly positive.
In order that $f(x_*) <0$ in some region, you then first need $$\log \left(\frac{b}{\sqrt{2a} }\right) < \frac 12$$ and this will be true up to the solution of $f(x)=0$ given in terms of Lambert function as Szeto already answered.
On
\begin{align} ax^2 &\le \ln(bx), \quad x,a,b>0 \tag{1}\label{1} . \end{align}
Considering limits, we have both \begin{align} \lim_{x\to0}ax^2 &> \lim_{x\to0}\ln(bx) ,\\ \lim_{x\to\infty}ax^2 &> \lim_{x\to\infty}\ln(bx) \quad \forall a,b>0 , \end{align}
thus there is no solution to \eqref{1} unless there is intersection point(s), which can be found in terms of Lambert $\operatorname{W}$ function, for example, as:
\begin{align}
ax^2 &= \ln(bx)
\tag{2}\label{2}
,\\
x^{-2}\ln(bx)&=a
,\\
-2x^{-2}\ln(bx)&=-2a
,\\
x^{-2}\ln((bx)^{-2})&=-2a
,\\
(bx)^{-2}\ln((bx)^{-2})&=-2ab^{-2}
,\\
\ln((bx)^{-2})&=\operatorname{W}(-2ab^{-2})
,\\
(bx)^{-2}&=\exp(\operatorname{W}(-2ab^{-2}))
,\\
x^{-2}&=b^2\exp(\operatorname{W}(-2ab^{-2}))
,\\
x&=(b^2\exp(\operatorname{W}(-2ab^{-2})))^{-\tfrac12}
,\\
x&=\frac1b\,\exp\left(-\tfrac12\operatorname{W}(-2ab^{-2})\right)
\tag{3}\label{3}.
\end{align}
Note that we ignore potential negative solution since we are interested in positive $x$.
The argument $-2ab^{-2}$ of
the Lambert $\operatorname{W}$ function in \eqref{3}
is negative, and from the well-known properties of
the Lambert $\operatorname{W}$ function,
we can deduce that there are only three possible cases,
depending on the values of $a,b$:
Case 1. There is no intersection for \begin{align} -2ab^{-2}&<-\exp(-1) ,\quad\text{that is, no solution to \eqref{1}} . \end{align}
Case 2. There are two intersections for \begin{align} -\exp(-1)&<-2ab^{-2}<0: \quad \begin{cases} x_1&=\frac1b\,\exp\left(-\tfrac12\operatorname{W}_0(-2ab^{-2})\right) ,\\ x_2&=\frac1b\,\exp\left(-\tfrac12\operatorname{W}_{-1}(-2ab^{-2})\right) , \end{cases} ,\quad\text{the solution to \eqref{1} is}\quad x\in[x_1,x_2] . \end{align}
Case 3. There is only one intersection (touch) for \begin{align} -2ab^{-2}&=-\exp(-1) ,\quad\text{the solution to \eqref{1} is just one value }\quad \\ x&=\frac1b\,\exp\left(-\tfrac12\operatorname{W}_0(-\exp(-1))\right) =\frac1b\,\exp(\tfrac12) . \end{align}
Illustration for $a=3$:
\begin{align} b_1&=3.5,\quad \text{no solution} ;\\ b_2&=6,\quad x\in[\tfrac16\exp(-\tfrac12 \operatorname{W}_0(-\tfrac16)), \tfrac16\exp(-\tfrac12 \operatorname{W}_{-1}(-\tfrac16))]) \approx[0.1846,0.6872] ;\\ b_3&=\sqrt{6\mathrm{e}}, \quad x=\tfrac{\sqrt6}6 \approx 0.4082 . \end{align}
On
Introducing $\epsilon$ we have
$$ a x^2=\ln b +\frac{1}{2}\ln x^2 +\epsilon^2 $$
and then
$$ e^{2ax^2}=2ax^2 \frac{b^2}{2ae^{2\epsilon^2}} \Rightarrow x = \sqrt{-\frac{1}{2a}W\left(-\frac{2ae^{2\epsilon^2}}{b^2}\right)} $$
but
$W\left(-\frac{2a}{b^2}\right) \le W\left(-\frac{2ae^{2\epsilon^2}}{b^2}\right) \Rightarrow x \ge \sqrt{-\frac{1}{2a}W\left(-\frac{2a}{b^2}\right)}$
Hint:
To solve $ax^2=\ln(bx)$:
$$2ax^2=\ln(b^2x^2)$$ Let $g=2ax^2$,
$$g=\ln(\frac{b^2}{2a}g)$$ $$e^g= \frac{b^2}{2a}g$$ $$\frac{-2a}{b^2}=-ge^{-g}$$ $$-g=W(\frac{-2a}{b^2})$$ $$x=\sqrt{\frac{-W(\frac{-2a}{b^2})}{2a}}$$
Now you can solve your inequality with suitably flipping the sign.