I am studying this non linear ODE:
$$x''=x-x^3+\nu x x'$$
I have already shown that the origin is a saddle point and that (1,0), (-1,0) are non-linear centers (they are centers of the linearization and the system is reversible).
Now I am asked to show that a necessary condition for a periodic solution of this equation is
$$\frac{1}{T} \int_0^T x^4 dt > \frac{1}{T} \int_0^T x^2 dt$$
Where $T$ is the period of the solution. I get that the idea to prove this must be that any periodic solution spends much more time in the region where $\vert x \vert > 1$ than in the region where $\vert x \vert < 1$ but I haven't been able to give a rigorous proof.
How can I prove this rigorously?
Hint: writing the equation in the form $$ x'=y,\quad y'=x-x^3+\nu x y, $$ we have $$ \left(-x^2+\frac12x^4+y^2\right)'=-2(x-x^3)x'+2yy'=2\nu xy^2. $$ So the best seems to be drawing the phase portrait for $\nu=0$ (without making linearizations!) and then deciding according to the sign of $2\nu xy^2$ where a periodic orbit can appear. The rest follows.