I've come across the following bound for $J_{\nu}$, the Bessel function of the first kind. It says that
$$\displaystyle |J^2_{\nu}(z)| \leqslant C_{\nu}|z|^{-1},$$
where $C_{\nu}$ is some constant depending on $\nu$. This would imply that $|J_{\nu}(z)| \leqslant C_{\nu}|z|^{-1/2}$. I've tried to prove this myself using the asymptotics of the Bessel function, which I found here. I found that
$$\displaystyle J_{\nu}(z) = -\sqrt{\frac{2}{\pi z}}\sin(z + a\nu) + O(z^{-3/2}),$$
where $a = -\frac{2\nu - 1}{4}.$ Doesn't this surely give us that $|J_{\nu}(z)| \leqslant C_{\nu}|z|^{-1/2} + C^{\prime}_{\nu}|z|^{-3/2}$? Why are we able to rid ourselves of the $|z|^{-3/2}$ term? Or is this derived in a completely different way altogether?
We easily get the bound if $|z| \geqslant 1$. But what about $0 < |z| < 1$? Is there some other asymptotic expansion I should be using for small $|z|$?