I have $r'=q/2$, $1<r<2$. From these I want to derive $$1+\frac{1}{(p/r)'}\ge r-1+\frac{1}{p/r}$$ and it is the same as $$3p'\le q.$$
I've tried several times, but I failed to derive second inequality from the first. And is there any relation between $r,r',p,p'$ and $(p/r)', (p/r)$? If so what is it?
I am sorry, I couldn't come up with a nice expression for $\left(\frac{p}{r}\right)'$. But I'm stubborn, so...
We have $\left(\frac{p}{r}\right)'=\frac{\frac{p}{r}}{\frac{p}{r}-1}=\frac{p}{p-r}$ and, since $r'=\frac{q}{2}$, we have $r=\frac{r'}{r'-1}=\frac{q}{q-2}$.
Then we have \begin{align} 1 + \frac{1}{\left(\frac{p}{r}\right)'} & \geq r-1 + \frac{1}{\left(\frac{p}{r}\right)} \\ 1 + \frac{p-r}{p} & \geq r-1 + \frac{r}{p} \\ 1+ \frac{p -\frac{q}{q-2} }{p} & \geq \frac{q}{q-2} -1 + \frac{\frac{q}{q-2}}{p} \\ 1 + \frac{pq -2p -q}{pq -2p} & \geq \frac{q}{q-2} -1 + \frac{q}{pq-2p} \\ \frac{pq -2p +pq-2p -q}{pq-2p} & \geq \frac{pq-pq +2p + q}{pq-2p} \\ 2pq-4p-q & \geq 2p + q \\ 2pq -2q & \geq 6p \\ pq -q & \geq 3p \\ q & \geq 3 \frac{p}{p-1}, \end{align} which is $q \geq 3p'$.
Note that we were able to simplify the denominator without changing the inequality, because by hypothesis $r<2$, hence $r'=\frac{q}{2} >2$ so that $q > 4$ and so $pq-2p>0$.