= Solution
At an <equilibrium point of a dynamical system>, the second equation gives
$$
x(s-xy)=0.
$$
The choice $x=0$ is impossible because then $\dot x=r>0$. Hence $y=s/x$, and substitution into the first equation gives
$$
0=r-(1+s)x+sx=r-x.
$$
Thus the unique fixed point is
$$
\boxed{(x_*,y_*)=\left(r,\frac sr\right)}.
$$
Choose
$$
\boxed{\alpha=\frac r{1+s}},
$$
which satisfies $0<\alpha<r$. Let
$$
R=\left\{(x,y):x\geq\alpha,\quad 0\leq y\leq\frac s\alpha,
\quad x+y\leq r+\frac s\alpha\right\}.
$$
We check the direction of the vector field on its four edges.
On $x=\alpha$,
$$
\dot x=r-(1+s)\alpha+\alpha^2y=\alpha^2y\geq0.
$$
On $y=0$,
$$
\dot y=sx>0.
$$
On $y=s/\alpha$, where $x\geq\alpha$,
$$
\dot y=sx\left(1-\frac x\alpha\right)\leq0.
$$
Finally, throughout the system,
$$
\dot x+\dot y=r-x.
$$
On the sloping edge $x+y=r+s/\alpha$, the bound $y\leq s/\alpha$ implies $x\geq r$, and hence $\dot x+\dot y\leq0$. Every boundary edge therefore points inward. Violations of the corresponding four inequalities are driven toward the boundary in the same order, so positive trajectories enter this compact <trapping region> and then remain there. This is the <Brusselator trapping region>.
The <Jacobian matrix> at the unique fixed point is
$$
J_*=
\begin{pmatrix}
-(1+s)+2x_*y_*&x_*^2\\
s-2x_*y_*&-x_*^2
\end{pmatrix}
=
\begin{pmatrix}
s-1&r^2\\
-s&-r^2
\end{pmatrix}.
$$
Its determinant and trace are
$$
\det J_*=r^2>0,
\qquad
\operatorname{tr}J_*=s-1-r^2.
$$
When $s-1>r^2$, the trace is positive, so <linear stability of a planar equilibrium> shows that the fixed point is a repeller. It lies in the interior of $R$.
Choose a trajectory in $R$ other than the equilibrium. Compactness of $R$ gives a nonempty compact omega-limit set, and the repelling equilibrium cannot belong to that set. Since there are no other equilibria, the <Poincaré-Bendixson theorem> gives a <periodic orbit>. Hence
$$
\boxed{s-1>r^2\quad\Longrightarrow\quad\text{the system has a periodic orbit}.}
$$
This is the <Brusselator periodic-orbit criterion>.
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