Use Poincaré-Bendixson to show that a limit cycle exists in the first quadrant. The entire problem reads:
Consider the predator-prey model $\dot x = \left( 4-x-\frac{2y}{1+x} \right)$, $\dot y = y(x-1)$. Assume that all positive solutions are bounded. (a) Find all critical points and determine their local stability. (b) Show that this system has a limit cycle in the first quadrant.
I have already found all of the equilibrium points, the important of which is an unstable focus at $(1,3)$. I attempted to show part (b) using some phase plane analysis, but I have hit a wall and I'm not sure where to go next. I started by plotting the parabola $y = \frac{(4-x)(1+x)}{2}$ because that's where $\dot x = 0$, but it didn't help me gain much. Any and all advice on how to do the rest of part (b) would be greatly appreciated!
 A: Let me repeat, with suitable modifications, my answer to the OP's earlier question.
Assume that all solutions starting in $\mathbb{R}^2_{+} := \{\, (x, y) : x \ge 0,\ y \ge 0 \,\}$ are bounded for $t > 0$ (this would require a separate proof). It follows then that for any such solution, its domain contains $[0,\infty)$ and its $\omega$-limit set is compact and nonempty.
Let $L$ stand for the $ω$-limit set of some point, $(x_0,y_0)$, sufficiently close to the unstable focus $(1,3)$. By the Poincaré–Bendixson theorem, as there are finitely many equilibria, $L$ is either a limit cycle, or a heteroclinic cycle, (EDIT: or a homoclinic loop), or an equilibrium.
There are two equilibria, $(4,0)$ and $(1,3)$. The first of them is an unstable focus, so it cannot belong to any heteroclinic cycle (because a heteroclinic cycle (EDIT: or a homoclinic loop) must contain an equilibrium that is an $\omega$-limit point for some other point).  Consequently, there are no heteroclinic cycles (EDIT: or homoclinic loops) at all.
So, $L$ is either a periodic orbit, or equals $\{(4,0)\}$.  We proceed now to excluding the latter.  The linearization of the vector field at $(4,0)$ has matrix
$$
\begin{bmatrix}
-1 & -\frac25
\\
0 & 3
\end{bmatrix}.
$$
There are two real eigenvalues, $-1$ and $3$, of opposite signs, so $(4,0)$ is a hyperbolic saddle.  Its stable manifold is tangent to an eigenvector corresponding to $-1$, that is, to $(1,0)$.  I claim that the stable manifold 
is the $x$-axis, minus $(4,0)$.  Indeed, on the $x$-axis we have $\dot{x} = 4 - x$, $\dot{y} \equiv 0$, so for any $(x_1,0)$ we have $\omega((x_1,0)) = \{(4,0)\}$.  Now, for a hyperbolic saddle its stable manifold is just the set of those points whose (unique) $\omega$-limit point is the saddle.  Hence, if $L = \{(4,0)\}$ then the positive semitrajectory of $(x_0, y_0)$ must belong to the $x$-axis, which contradicts the uniqueness of the initial value problem (notice that $y_0 > 0$).
We have thus shown that $L$ is a periodic orbit.
