Suppose I have the following function $f: \mathbb{R} \to \mathbb{R}$: $$f(r) = (\gamma^2 -\epsilon^2) - 2 r^2 - r \frac{(\gamma -3\epsilon)\phi\left(\frac{\epsilon-\gamma}{r}\right) - (\gamma+\epsilon)\phi\left(-\frac{\gamma+\epsilon}{r}\right) }{\Phi\left(\frac{\epsilon-\gamma}{r}\right)-\Phi\left(-\frac{\gamma+\epsilon}{r}\right)}$$
where $\phi$ and $\Phi$ are respectively the standard gaussian pdf and cdf, and $\gamma>\epsilon>\frac{2}{3}\gamma>0$ are real-valued constants.
I am looking for a statement along the lines of: if $r>g(\epsilon,\gamma)$ then $f(r)<0$.
Here is a plot of how the function looks like for some fixed value of $\gamma$ and $\epsilon$.
Unfortunately, I was only able to show that $f(r) \overset{r \to \infty}{\longrightarrow} \epsilon (\gamma - \frac{5}{3}\epsilon)$ which is a weaker statement. Any idea on how to tackle this problem is welcome (e.g. a tight upper-bound on the ratio of gaussian pdf to cdf).
(EDIT) It might be helpful to know where does $f(r)$ comes from. I was originally interested in showing that the following quantity is negative: $$(\gamma^2 -\epsilon^2) - (\gamma+\epsilon) \mathbb{E}[X] - 2\mathbb{E}[X^2 ] $$ where $X \sim \mathcal{N}(0,r^2)$ is a gaussian r.v. truncated on $(\alpha,\beta)$ with $\alpha = -(\gamma+\epsilon)$, $\beta = -(\gamma-\epsilon)$. Plugging in the known moments of the truncated gaussian we get $f(r)$.
(UPDATE) I was able to find a point $r_0=10(\gamma+\epsilon)$ such that $f(r_0) < 0$. Hence it would be enough to show that $f$ monotonically decreases for $r>r_0$ (numerically this seems to be the case).