For any given probability density function $p$ (with finite second moments) on $\mathbb{R}^d$ I want to show that the following integral is bigger than $\textit{(or equal to)}$ zero
$$ \int_{\mathbb{R}^d} \Big(\Delta\Psi(x)-\|\nabla \Psi(x)\|^2\Big) p(x)dx. $$
Obviously here I need to impose some assumptions on the function $\Psi$.
Since this needs to hold for any probability density $p$, I think the easiest thing to ask is :
Are there any conditions I can impose on a function $\Psi :\mathbb{R}^d \to \mathbb{R}$ such that
$$ \Delta\Psi(x)-\|\nabla \Psi(x)\|^2 \geq 0, ~~\text{for}~a.e~x\in\mathbb{R}^d. $$
(Let me work in $d=1$) If you truly want our integral to be "bigger than zero", i.e. strictly positive, then your condition isn't enough for that. What would work would be something like $\partial_{xx}\psi-(\partial_x \psi)^2 \ge \gamma$ for all $x$, for some $\gamma >0$. You can notice that this last condition is equivalent to $-\partial_{xx}(e^{-\psi})\ge e^{-\psi}\gamma$ as a side note.