At some point in this article, the following lemma is applied.
Lemma
If $u\in H^1(\alpha,\beta),u(\beta)=0,\beta>\alpha\geq0$, then:
$$\int_\alpha^\beta|u(r)|^2rdr\leq2^{-1}(\beta-\alpha)^2\int_\alpha^\beta|u'(r)|^2rdr.$$
I tried three different routes, and got three different inequalities, each of which differs from the above. Here goes.
Route 1
\begin{align*} \int_\alpha^\beta|u(r)|^2rdr\overset\ast={}&\int_\alpha^\beta\left|\int_\beta^ru'(s)ds\right|^2rdr\overset{J}{\leq}\int_\alpha^\beta\int_r^\beta|u'(s)|^2rdsdr\leq{} \\ {}\leq{}&\int_\alpha^\beta\int_\alpha^\beta|u'(s)|^2rdsdr=\int_\alpha^\beta|u'(s)|^2ds\cdot\int_\alpha^\beta rdr={} \\ {}={}&\frac12(\beta^2-\alpha^2)\int_\alpha^\beta|u'(s)|^2ds. \end{align*}
J is Jensen's inequality, $\ast$ is because $H^1$ functions in 1 dimension are absolutely continuous so $u(r)-u(\beta)=\int_\beta^ru'(s)ds$ and $u(\beta)=0$. Problem: $r$ disappeared from the integrand.
Route 2
This is a modification of the above which manages to keep the $r$ (or rather, the $s$), but the constant is still not the right one.
\begin{align*} \int_\alpha^\beta|u(r)|^2rdr\overset\ast={}&\int_\alpha^\beta\left|\int_\beta^ru'(s)ds\right|^2rdr\overset{J}{\leq}\int_\alpha^\beta\int_r^\beta|u'(s)|^2rdsdr\leq{} \\ {}\overset\star\leq{}&\int_\alpha^\beta\int_r^\beta|u'(s)|^2sdsdr\leq\int_\alpha^\beta\int_\alpha^\beta|u'(s)|^2sdsdr={} \\ {}={}&\int_\alpha^\beta|u'(s)|^2sds\cdot\int_\alpha^\beta 1dr={} \\ {}={}&(\beta-\alpha)\int_\alpha^\beta|u'(s)|^2sds. \end{align*}
Basically I just figured out I could end up with the right integrand if I exploited the fact that we are integrating where $r\leq s$, which is what justifies $\star$, while J and $\ast$ are the same as above.
Route 3
\begin{align*} \int_\alpha^\beta|u(r)|^2rdr={}&\int_\alpha^\beta|u(r)\sqrt r|^2dr\overset\ast=\int_\alpha^\beta\left|\int_\beta^r\frac{d}{dr}(u\sqrt r)|_s\right|^2dr\leq{} \\ {}\overset{J}{\leq}{}&\int_\alpha^\beta\int_r^\beta\left|u'(s)\sqrt s+\frac{u(s)}{\sqrt s}\right|^2dsdr=\int_\alpha^\beta\int_r^\beta\left|\frac{2u'(s)s+u(s)}{2\sqrt s}\right|^2dsdr\leq{} \\ {}\overset\ast\leq{}&\int_\alpha^\beta\int_r^\beta\frac12\left(\left|\frac{u'(s)s}{\sqrt s}\right|^2+\left|\frac{u(s)}{\sqrt s}\right|^2\right)dsdr\leq{} \\ {}\leq{}&\int_\alpha^\beta\frac12\left(|u'(s)|^2s+\frac{|u(s)|^2}{s}\right)dsdr={} \\ {}={}&\frac12(\beta-\alpha)\int_\alpha^\beta\left(|u'(s)|^2+\frac{|u(s)|^2}{s}\right)ds. \end{align*}
J is as above, $\ast$ is as above ($u(r)\sqrt r$ is Sobolev because $r$ is smooth and $u$ is Sobolev), $\star$ is the convexity of squaring.
How do I get the article's inequality though?
Update
As pointed out by @MichałMiśkiewicz in the comments, I made a mistake in all three routes: Jensen's inequality only works for probability spaces, so I have to have an integral over a probability space in order to apply it. Here is the correct version of Route 2:
\begin{align*} \int_\alpha^\beta|u(r)|^2rdr\overset\ast={}&\int_\alpha^\beta\left|\int_\beta^ru'(s)ds\right|^2rdr={} \\ {}={}&\int_\alpha^\beta\left|(\beta-\alpha)\frac{1}{\beta-\alpha}\int_\beta^ru'(s)ds\right|^2rdr\leq{} \\ {}\overset{J}{\leq}{}&\int_\alpha^\beta(\beta-\alpha)^2\left(\frac{1}{\beta-\alpha}\int_r^\beta|u'(s)|^2rds\right)dr\leq{} \\ {}\overset\star\leq{}&(\beta-\alpha)\int_\alpha^\beta\int_r^\beta|u'(s)|^2sdsdr\leq{} \\ {}\leq{}&(\beta-\alpha)\int_\alpha^\beta\int_\alpha^\beta|u'(s)|^2sdsdr={} \\ {}={}&(\beta-\alpha)\int_\alpha^\beta|u'(s)|^2sds\cdot\int_\alpha^\beta 1dr={} \\ {}={}&(\beta-\alpha)^2\int_\alpha^\beta|u'(s)|^2sds. \end{align*}
In general, what changes is I get an extra factor of $\beta-\alpha$. However, this gets me a missing $\frac12$ in Route 2, and an extra $\beta+\alpha$ I would like to get rid of from Route 1. So how do I solve this?
As @mickep pointed out, I have tu multiply and divide by $\beta-r$, not $\beta-\alpha$. So finally here is the proof of that inequality:
\begin{align*} \int_\alpha^\beta|u(r)|^2rdr\overset\ast={}&\int_\alpha^\beta\left|\int_\beta^ru'(s)ds\right|^2rdr={} \\ {}={}&\int_\alpha^\beta\left|(\beta-r)\frac{1}{\beta-r}\int_\beta^ru'(s)ds\right|^2rdr\leq{} \\ {}\overset{J}{\leq}{}&\int_\alpha^\beta(\beta-r)^2\left(\frac{1}{\beta-r}\int_r^\beta|u'(s)|^2rds\right)dr\leq{} \\ {}\overset\star\leq{}&\int_\alpha^\beta(\beta-r)\int_r^\beta|u'(s)|^2sdsdr\leq{} \\ {}\leq{}&\int_\alpha^\beta(\beta-r)\int_\alpha^\beta|u'(s)|^2sdsdr={} \\ {}={}&\int_\alpha^\beta|u'(s)|^2sds\cdot\int_\alpha^\beta(\beta-r)dr={} \\ {}={}&-\frac12(\beta-r)^2|_r^\beta\int_\alpha^\beta|u'(s)|^2sds={} \\ {}={}&\frac12(\beta-\alpha)^2\int_\alpha^\beta|u'(s)|^2sds, \end{align*}
just what we wanted. QED