Show that the system, $$x'=-x-y+x(x^2+2y^2)$$ $$y'=x-y+y(x^2+2y^2)$$ has at least one peridic solution.
I know that I need to use the Poincare Bendixon Theorem, but I'm not to sure how to find the trapping region. When my teacher did an example in class he basically did a proof by picture and made it seem like all arrows were pointed inward within the trapping region. I'm wondering how would I find and rigorously prove that a trapping region really has all arrows pointed inward? Any help is appreciated, thanks!
EDIT: I believe I have figured out the trapping region to be the ellipse $x^2+2y^2=1$ and I believe I have proven it by looking at cases depending on which quadrant the coordinates is in.
Now my question is, how do I deal with the fixed point $(0,0)$ within the trapping region?
Because of the form of the linear part of the vector field it seems advisable to explore the dynamic of the Euclidean radius. For simplicity of computation, use $E=\frac12r^2=\frac12(x^2+y^2)$ to get $$ \frac{d}{dt}E=x\dot x+y\dot y=-2E+2E(x^2+2y^2) $$ so that $$ 2E(2E-1)\le\dot E\le 2E(4E-1) $$ This means that $\dot E$ is negative for $0< E< \frac14$ and positive for $E>\frac12$. Looking in the direction of negative times, the time reversed ODE, this means that the annulus $\frac14<E<\frac12$, $\sqrt{\frac12}<r<1$, is invariant in that time direction and thus has to contain a limit cycle which is also a periodic solution.