= Solution
Let
$$
F(N)=\frac{rN}{(1+bN)^2}.
$$
At low density, $F(N)/N\to r$, so $r$ is the density-independent per-generation multiplication factor. The parameter $b$ has units of inverse population and sets the density scale at which crowding suppresses recruitment.
The <fixed points> satisfy $N=F(N)$. Besides extinction $N=0$, for $r>1$ there is the positive fixed point
$$
\boxed{N_*=\frac{\sqrt r-1}{b}.}
$$
Since
$$
F'(N)=\frac{r(1-bN)}{(1+bN)^3},
$$
its multiplier is
$$
F'(N_*)=\frac{2-\sqrt r}{\sqrt r}=\frac2{\sqrt r}-1.
$$
For every $r>1$, $-1<F'(N_*)<1$, so <fixed point stability for an iteration> proves that $N_*$ is \b[locally asymptotically stable].
Scale the population by $x_\tau=bN_\tau$. Then
$$
x_{\tau+1}=f(x_\tau),\qquad f(x)=\frac{rx}{(1+x)^2}.
$$
The map increases on $0<x<1$, decreases on $x>1$, and has global maximum $f(1)=r/4$. Starting from $x_1=1$ therefore gives
$$
x_2=\frac r4=:U,
\qquad
x_3=f(U)=\frac{4r^2}{(4+r)^2}=:L.
$$
For $r>4$, writing $s=\sqrt r>2$ gives
$$
L-1=\frac{(s-2)(s+2)(3s^2+4)}{(s^2+4)^2}>0,
\qquad
(s-1)-L=\frac{(s-2)^3(s^2+s+2)}{(s^2+4)^2}>0.
$$
Thus $1<L<N_*b=s-1<U$. Since $f$ decreases on $[L,U]$, $f(U)=L$ and $f(L)>f(s-1)=s-1>L$, while the global maximum gives $f(L)\leq U$. Hence $f([L,U])\subseteq[L,U]$. The <cobweb plot> consequently remains in this invariant interval after the first iterate, and both endpoints are attained at $\tau=2,3$:
$$
\boxed{\frac{4r^2}{(4+r)^2b}\leq N_\tau\leq\frac r{4b}\qquad(\tau\geq2).}
$$
The printed bound cannot include $\tau=1$, because $N_1=1/b<L/b$ when $r>4$.
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