Let $p(x)$ and $q(x)$ be two smooth probability density functions defined on the whole real line s.t. $p(x),q(x) > 0 \ \forall x \in R$.
Do there exist conditions that one can impose on $p(x)$ and $q(x)$ and the relation between them so that
$$\int_{-\infty}^{+\infty} \left( \frac{p(x)}{q(x)}-1 \right) q'(x)^2 \, dx < 0$$ ?
Is there a way to approach this problem maybe through the calculus of variations minimizing a functional?
Too long for a comment. The idea is that you can construct classes of $p,q$ so that your inequality is true.
If you let $p(x)=q(x)(1+r(x))$, where $|r(x)|<1$; then you need two conditions to be satisfied: normalization of $p$: $$ \int_{-\infty}^\infty q(x)r(x)dx=0, $$ and your inequality, $$ \int_{-\infty}^\infty r(x)q'(x)^2dx<0. $$
Let us suppose for simplicity that $r$ and $q$ are even functions, so that it suffices to show $$ \int_0^\infty q(x)r(x)dx=0,\\ \int_0^\infty r(x)q'(x)^2dx<0. $$ Let us also suppose that $q=Q$ (constant) on $I=[0,1]$, and decreases to $0$ on $[1,2]$, with $q=0$ on $[2,\infty)$. Then we need $$ Q\int_0^1r(x)dx+\int_1^2q(x)r(x)dx=0,\\ \int_1^2r(x)q'(x)^2dx<0. $$ You can then choose $r<0$ on $[1,2]$ and $r>0$ on $[0,1]$ appropriately to get a class of solutions.