Given the equation $$x''=-x+\frac{1}{6} x^3 $$ Prove that it has periodic solutions in the neighborhood of $x=0$.
I am given the proof of this which is the following.
First it's a basic exercise to find the energy $$E=(x')^2+x^2-\frac{1}{12} x^4 $$
For small $E$, consider $$x'=\pm \sqrt{ E-x^2+\frac{1}{12} x^4 }$$
For small $E$, $x \sim \sqrt{E} $, thus there are periodic solutions in a neighborhood of $0$.
I don't understand the last part, first of all why is it $x \sim \sqrt{E} $ for small $E$ and second how does this prove that the equation has periodic solutions around $0$?
As said by @Sal, by solving $E=V(x)$, where $V(x)=x^2-x^4/12$ find four real turning points $\pm x_1, \pm x_2$ $|x_1|<|x_2|$. it turns out that for $E\in (0,3)$ these are real, then for each $E$ in this interval plot $p(x)=\pm\sqrt{E-V(x)}, x \in [-x_1,x_1]$ to see the closed phase-space orbits which indicate periodic motion between the innermost turning points. For example see the phase space $(x,p(x))$ plot for $E=2, x_1=1.59.$ below.