In 3 dimensions, does it hold that $\int r^{-2} h\le C\int r^{-1}(1+r)^{-1} h$?

Question

in theorem 9 from these notes notes, after computing the divergence of the flux, I get the coefficient of the $\phi^2$ term to be similar to $\frac 1{r(1+r)^{2+\delta}}$, so upon plugging into the integrand, I would actually get something of the form $$\int_D \langle r\rangle^{-1-\delta}\left(|\partial \phi|^2 + \frac 1{r(1+r)}\phi^2\right), $$ yet in the notes, I see $\frac 1{r(1+r)^{2+\delta}}$ replaced by $r^{-2}$. At first I thought this might follow from some inequality like $$ \int r^{-2} h\le C\int r^{-1}(1+r)^{-1} h, $$ but the more I think about it, the less this inequality makes sense. I was thinking that maybe, since $r^{-a}$ is integrable near the origin for $a<3$ that I could reasonably make such a substitution (the inequality holds for constants, but I couldnt find a continuity argument to extend the result for general $h$). At infinity there are no problems, but close to $0$ one could find functions whose mass is canceled out by the vanishing factor of $r$ that survives after writing the integrals in polar coordinates. Is there something I am missing?