Find all monotonic functions $f : \mathbb{R}\to\mathbb{R}$ that satisfy $f''(x) + xf'(x) - (\delta+\gamma f(x))(f'(x))^2>0$ I would like to find a set of functions that satisfy the following:

Find all monotonic functions $f : \mathbb{R}\to\mathbb{R}$ that satisfy $f''(x) + xf'(x) - (\delta+\gamma f(x))(f'(x))^2>0$

where $\gamma$ and $\delta$ are finite, non-zero constants. (A previous version of the question had $\gamma = \delta \equiv 1$, but this need not necessarily be the case).
The equation in the inequality is the derivative of a generalised log-odds function, the condition implying the final logistic curve is strictly increasing. Ideally, bounding the inequality by taking limits as $x\to\pm\infty$ should lead to an expression dependent solely on the parameters $\gamma$ and $\delta$ ${}^{1}$.
Properties that would naturally seem to be needed are $f'(x)\to 0$ and $f''(x)\to 0$ as $x \pm\infty$. Two good candidates I have found that would satisfy this are:

*

*$f(x) = \text{arcsinh}(x)$, and

*$f(x) = \tanh{(x)}$,

yet I'm not sure how to exploit their limiting behaviour to reduce the inequality to a simple form for $\gamma$ and $\delta$ (i.e., no explicit dependence on $x$). How can I do this in a rigorous manner? Are there any other suitable monotonic functions $f : \mathbb{R} \to \mathbb{R}$?

${}^{1}$ This is for optimisation purposes guaranteeing the monotonicity of the logistic curve for some functional form of $f(x)$. In reality $x$ will never reach $\pm\infty$; it will exist in the interval $\left(\Phi^{-1}(\varepsilon),\Phi^{-1}(1-\varepsilon)\right)$ where $\mathcal{O}(\varepsilon) \sim 10^{-6}$. Therefore, for practical purposes, if bounding the inequality via the asymptotic limiting behaviour of $f(x)$ is too troublesome, we can also bound it by assessing its value as $x\to \pm 5$.
 A: Not an answer, just some ideas.
The inequality can be rewritten, when $f'(x)\neq0,$ as:
$$f'(x)\left(\frac{f''(x)}{f'(x)}+x-(f(x)+1)f'(x)\right)>0\tag1$$
The interior is the derivative of $$g(x)=\ln|f'(x)|+\frac{x^2-(f(x)+1)^2}2,$$ so $(1)$ means when $f(x)$ is increasing,  then $g(x)$ is increasing and when $f(x)$ is decreasing then $g(x)$ is decreasing.
If $f'(x)=0,$ then we see $f''(x)>0$ so $f(x)$ is a local minimum. Thus, there can be at most one such point.
It seems very difficult to make $g(x)$ increasing if $f(x)$ increases faster than $x.$ That's because you'll need $\ln|f'(x)|$ to compensate for the increases in $f^2(x).$
It seems more likely to find examples with $f'(x)\to 0.$

An example which does work is $f(x)=\log(1+x^2)-1.$
Then $f'(x)=\frac{2x}{x^2+1},$ $f''(x)=\frac{2-2x^2}{(1+x^2)^2}$ and you get:
$$f''(x)+xf'(x) -(f(x)+1)(f'(x))^2=\frac{2+2x^4-4x^2\log(1+x^2)}{(1+x^2)^2}$$
Wolfram Alpha gives this is positive. In fact, there are smaller $c$ such that $f(x)=\log(1+x^2)-c$ works.
Specifically, you need:
$$\frac{1+x^4}{2x^2}-\log(1+x^2)>1-c$$
for all $x\neq 0.$ looks like the infimum of possible $c$ is $\approx 0.5941.$
Another example is $f(x)=\frac{x^2}{x^2+1}-c,$ for $c>2.$ I don't have the minimum there.
It seems like a lot of functions $f$ with $f'\to 0$ as $x\to\pm \infty$ with have some value $C$ such that $f(x)-c$ will work for $c$ large enough. You'll definitely need $f$ to have at most one local minimum and no other points where $f'=0.$
I haven't found any strictly increasing $f$ which works yet.
