= Solution
The physical <state space> is the <positively invariant set> $X,Y,Z\geq0$, $X+Y+Z=N$: adding the equations gives zero, and the vector field points inward at each coordinate boundary. At an <equilibrium point>, either $Y=0$, giving $(X,Y,Z)=(N,0,0)$, or $\beta X=b+r$. Therefore
$$
\boxed{N_c=\frac{b+r}{\beta},\qquad X_s=N_c,\quad Y_s=\frac{b(N-N_c)}{b+r},\quad Z_s=\frac{r(N-N_c)}{b+r}.}
$$
A positive endemic equilibrium exists exactly when $N>N_c$.
For $N<N_c$, put $d=b+r-\beta N>0$. Since $X\leq N$,
$$
\dot Y\leq-dY,\quad 0\leq Y(t)\leq Y(0)e^{-dt},\qquad
Z(t)=Z(0)e^{-bt}+r\int_0^t e^{-b(t-s)}Y(s)\,ds\longrightarrow0.
$$
The last limit follows either by evaluating the exponential integral (with a $te^{-bt}$ term when $d=b$), or by splitting it into an early decaying part and a late part with uniformly small $Y$. Hence $X\to N$ for every physical initial condition. Although immune numbers need not decrease monotonically initially, both infected and immune numbers tend to zero.
At the endemic equilibrium, the <Jacobian matrix> of the $X,Y$ subsystem is
$$
J=\begin{pmatrix}-b-\beta Y_s&-\beta X_s\\ \beta Y_s&0\end{pmatrix}.
$$
A disturbance proportional to $e^{pt}$ therefore obeys
$$
\boxed{p^2+(b+\beta Y_s)p+\beta^2X_sY_s=0.}
$$
Both coefficients are positive. Real roots have negative sum and positive product; complex conjugate roots have real part $-(b+\beta Y_s)/2<0$. Thus the endemic equilibrium is <asymptotically stable> by <linear stability analysis>.
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