This is not an answer, just a comment following k3thomps computations in remarks, the question is equivalent to, it seems: why is
$$
\int_0^\infty \left(\left(\frac{n^2}{x^2}-1\right) J_n^2 +(J_n^\prime)^2\right)J_0^2\text{d}x> 0 \,?
$$
Below is an alternative formulation of the same question (I can't make it work, but here it is anyway).
We have, by definition, $$J_n^{\prime\prime}+\frac{1}{x}J_n ^\prime-\left(\frac{n^2}{x^2}-1\right) J_n=0$$
Integrating against $J_nJ_0^2$, $n\geq1$, we obtain
\begin{eqnarray*}
&&\int_0^\infty \left(\left(\frac{n^2}{x^2}-1\right) J_n^2 + (J_n^\prime)^2\right) J_0^2 \text{d}x\\
&=& \frac{1}{2}\int_0^\infty -(J_n^2)^\prime(J_0^2)^\prime +\frac{1}{x}(J_n^2)^\prime J_0^2\text{d}x\\
&=& \frac{1}{2}\int_0^\infty J_n^2\left((J_0^2)^{\prime\prime} -\left(\frac{1}{x}J_0^2\right)^\prime \right)\text{d}x.
\end{eqnarray*}

This is puzzling, because $$(J_0^2)^{\prime\prime} -\left(\frac{1}{x}J_0^2\right)^\prime$$ oscillates on $(0,\infty)$.

Nevertheless, using the idea suggested in the original post, let $Q_{n}\left(r\right)=J_{n}^{2}\left(r\right).$ We have
\begin{align*}
\int_{0}^{\infty}\left(\frac{1}{r}J_{n}^{\prime}J_{n}+\left(J_{n}J_{n}^{\prime}\left(r\right)\right)^{\prime}\right)J_{0}^{2}\text{d}r & =\frac{1}{2}\int_{0}^{\infty}\left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right)Q_{0}\text{d}r\\
& =\frac{1}{2}\int_{0}^{\infty}\left(-\frac{1}{r}Q_{0}+Q_{0}^{\prime}\right)^{\prime}Q_{n}\text{d}r.
\end{align*}
Now,
\begin{align*}
Q_{0}\left(r\right) & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\psi\right)\text{d}{\psi},\\
Q_{n}\left(r\right) & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi},
\end{align*}
and
\begin{align*}
Q_{n}^{\prime} & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime}\left(2r\sin\varphi\right)2\sin\varphi\cos\left(2n\varphi\right)\text{d}{\varphi}\\
Q_{n}^{\prime\prime} & =\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime\prime}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)^{2}\cos\left(2n\varphi\right)\text{d}{\varphi}
\end{align*}
Recall that
$$
rJ_{0}^{\prime\prime}(r)+J_{0}^{\prime}(r)+rJ_{0}=0,
$$
thus
\begin{align*}
Q_{n}^{\prime\prime}= & \frac{1}{r}\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime\prime}\left(2r\sin\varphi\right)\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi}\\
= & -\frac{1}{r}\frac{1}{\pi}\int_{0}^{\pi}J_{0}^{\prime}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi}\\
& -\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi},\\
\left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right) & =-\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)\left(2\sin\varphi\right)\cos\left(2n\varphi\right)\text{d}{\varphi},\\
\left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right)Q_{0} & =-\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\varphi\right)2\sin\varphi\cos\left(2n\varphi\right)\text{d}{\varphi} \times \\
&\frac{1}{\pi}\int_{0}^{\pi}J_{0}\left(2r\sin\theta\right)d\theta,\\
\frac{1}{2}\int_{0}^{\infty}\left(\frac{1}{r}Q_{n}^{\prime}+Q_{n}^{\prime\prime}\right)Q_{0}\text{d}r & =-\frac{1}{\pi^{2}}\int_{0}^{\pi}\text{d}\theta\int_{0}^{\pi}2\sin\varphi\cos\left(2n\varphi\right)\text{d}{\varphi} \times \\
&\int_{0}^{\infty}J_{0}\left(2t\sin\theta\right)J_{0}\left(2t\sin\varphi\right)\text{d}t
\end{align*}
Now, it appears that for $\alpha,\beta>0$ $\int_{0}^\infty J_{0}(\alpha r)J_{0}(\beta r) \text{d} r $ is a known quantity, a so-called complete elliptic integral of the first type named $K$, given by
$$
\int_{0}^\infty J_{0}(\alpha r)J_{0}(\beta r) \text{d} r = \frac{2}{\pi \max(\alpha,\beta)} K \left(\frac{\min(\alpha,\beta)}{\max(\alpha,\beta)}\right)
$$

Now, more DLMF or algebraic manipulations to sort out the subsequent integrals on triangles..to be continued (one day).