Approximating monotonically increasing differential equation I am trying to make sense of the Appendix of the paper (Cooper, 1986).
The following model is presented:
$$\dot{(BX)}=\gamma_1BX \\
\dot{(BXB)}=\gamma_2(BX)B \\
\dot{B}=\gamma_3(BXB)$$
Without assuming for now anything about $\dot{X}$, namely how it relates to $B(t)$, he makes the following remark: $\dot{B}=\gamma_B\int_0^tB(t')Z(t')dt'$, where $Z(t)=\int_0^tX(t')B(t')dt'$ and $\gamma_B=\gamma_1\gamma_2\gamma_3$.
So far so good. What I cannot make sense is the following:

Since $B$ is monotonically increasing, $\dot{B} \leq \gamma_BB^2\int_0^tdt''\int_0^{t''}dt'\times X(t')$. It follows that $B(t) \leq B_0[1-\gamma_BB_0\int_0^tdt'''\int_0^{t'''}dt''\int_0^{t''}dt'\times X(t')]^{-1}$.

To me -- this, and everything that follows it -- seem abacadabra. If anyone can illuminate the cited text, I would be ecstatic! :)
 A: We have
\begin{align*}
  B'(t) &= \gamma_B\int_0^t  B(t')\int_0^{t'} X(t'')B(t'')\, dt''\, dt' 
\end{align*}
Since $B$ is increasing, for $t'' \le t' \le t$ we have $B(t'') \le B(t') \le B(t)$, this gives, as $X \ge 0$, that
\begin{align*}
  B'(t) &= \gamma_B\int_0^t  B(t')\int_0^{t'} X(t'')B(t'')\, dt''\, dt'\\
        &\le gamma_B \int_0^t B(t)\int_0^{t'} X(t'')B(t)\, dt''\, dt'\\
        &= \gamma_B B^2(t) \int_0^t \int_0^{t'} X(t'')\,dt''\, dt'\\
        &= \gamma_B B^2(t) \int_0^t \int_0^{t''} X(t')\,dt'\, dt''\\
\end{align*}
Let $Y(t) := \gamma_B\int_0^t \int_0^{t''} X(t')\,dt'\, dt''$, then $B' \le Y\cdot B^2$ for all $t$. We get
$$ \int_0^{t} \frac{B'}{B^2}\, dt \le \int_0^t Y(t)\, dt $$
Hence
$$ \frac 1{B(0)} - \frac 1{B(t)} \le \int_0^t Y(t)\, dt $$
or 
$$ \frac 1{B(t)} \ge \frac 1{B_0} - \int_0^t Y(t)\, dt \iff
  B(t) \le \frac{B_0}{1 - B_0\int_0^t Y(t)\, dt}.
$$
Now plug $Y$ into the last equation, giving
$$ B(t) \le \frac{B_0}{1 - B_0\gamma_B\int_0^{t}\int_0^{t'''}\int_0^{t''} X(t')\,dt'\, dt''\, dt'''} $$
