Showing that $z^2 e^{-z^2/2} \int \frac{\phi^2(x)}{\cosh(xz)} \, dx \geq \frac{1}{2\sqrt{\pi}} \frac{z^2}{z^2+1}$

Question

Let $\phi(x) = \frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$ denote the standard Normal density. We define the function $$ f(z) = z^2 e^{-z^2/2} \int \frac{\phi^2(x)}{\cosh(xz)} \, dx. $$ Note that we also have, if $X$ is standard Normal, $$ f(z) = z^2 \exp(-z^2/2) \,\mathbb{E}_X\Big[\frac{\phi(X)}{\cosh(zX)}\Big]. $$

Desired inequality: I would like to show that $$ f(z) \geq \frac{z^2}{z^2 + 1} \frac{1}{2 \sqrt{\pi}}, \quad \mbox{for all}~z > 0. $$

So far, I have checked that $f(0) = f'(0) = 0$ and therefore by Taylor's theorem we have that $$ f(z) \geq \frac{z^2}{2} \inf_{0 \leq \xi \leq z} f''(\xi). $$ Thus, it would be enough to show that $$ f''(\xi)(1 + \xi^2)\geq \frac{1}{\sqrt{\pi}} \quad \mbox{for any}~\xi > 0. $$ For this, note that the right hand side is actually an equality at $\xi = 0$, so it is enough to show that $f''(\xi)(1 + \xi^2)$ is an increasing function. Numerically, this appears to be true, but I am having difficulty showing it.

Answer 1

In equivalent terms, we are asked to prove that $$ g(z)=(z^2+1)e^{-z^2/2}\int_{\mathbb{R}}\frac{e^{-x^2}}{\cosh(xz)}\,dx \geq \frac{1}{\sqrt{2}} $$ but this cannot be true for all $z\in\mathbb{R}$, since $g$ is a Schwartz function, so it has to decay to zero (and pretty fast, too) as $|z|\to +\infty$. Anyway, let us find some sensible lower bounds for $$ h(z)=\int_{\mathbb{R}}\frac{e^{-x^2}}{\cosh(xz)}\,dx =\frac{\sqrt{\pi}}{|z|}\int_{\mathbb{R}}\frac{e^{-x^2}}{\cosh(\pi x/z)}\,dx$$ where the equality follows by considering the Fourier transforms of $e^{-x^2}$ and $\text{sech}(x)$.
Obviously $h$ is an even function, and by the equality above it fulfills $$ h(z) = \frac{\sqrt{\pi}}{z} h(\pi/z) $$ for any $z>0$. It follows that the best global lower bounds we can get are of the form $$ \left\{\begin{array}{rcl}h(z)\geq p(|z|)&\text{ if }&|z|\leq \sqrt{\pi}\\h(z)\geq\frac{\sqrt{\pi}}{|z|}\,p\left(\frac{\pi}{|x|}\right)&\text{ if }&|z|\geq \sqrt{\pi}\end{array}\right. $$ like $$ h(z) \geq \frac{C}{\sqrt{\pi+z^2}},\quad C\approx 1.486 $$ Always by the reflection formula $$ h(z)\sim\frac{\sqrt{\pi}}{z} $$ as $z\to +\infty$.