<s>[INST] Non-empty limit set for the dynamical system : $x_1' = x_1 + 2x_2 - 2x_1(x_1^2 + x_2^2)^2, \space x_2' = 4x_1 + 3x_2 - 3x_2(x_1^2 + x_2^2)^2 $ 
Using the Lyapunov Function $V=\frac{1}{2}x_1^2 + \frac{1}{2}x_2^2$ , prove that the omega(ω)-limit set is non-empty for any initial value given for the dynamical system :
  $$x_1' = x_1 + 2x_2 - 2x_1(x_1^2 + x_2^2)^2$$
  $$x_2' = 4x_1 + 3x_2 - 3x_2(x_1^2 + x_2^2)^2$$

Attempt : 
$$\nabla V(x)F(x)=x_1(x_1 + 2x_2 - 2x_1(x_1^2+x_2^2)) + x_2(4x_1+3x_2-3x_2(x_1^2+x_2^2))$$
$$=$$
$$x_1^2 + 2x_2x_1 - 2x_1^2(x_1^2+x_2^2)^2 + 4x_1x_2 + 3x_2^2 -  3x_2^2(x_1^2 + x_2^2)^2$$
$$=$$
$$x_1^2 + 3x_2^2 + 6x_1x_2 - 2x_1^2(x_1^2 + x_2^2)^2 - 3x_2^2(x_1^2+x_2^2)^2$$
$$\leq$$
$$3x_1^2 + 3x_2^2 + 6x_1x_2 - 2x_1^2(x_1^2 + x_2^2)^2-2x_2^2(x_1^2 + x_2^2)$$
$$=$$
$$3(x_1+x_2)^2 - 2(x_1^2 + x_2^2)^2(x_1^2 + x_2^2)$$
$$\leq$$
$$3(x_1+x_2)^2 - (x_1^2 + x_2^2)^3 = 3(x_1+x_2)^2 - V^3(x)$$
Now, I don't know how to continue from this point forward. I know I have to make the final expression in terms of $V(x)$ but I can't see how I would convert $3(x_1+x_2)^2$ to express it in terms of $V(x)$. 
If I have any mistake or if I could go on another approach regarding the inequalities, I would really appreciate some help for these or for what I should do next.
 [/INST] The Cauchy's inequality can be rewriten in the form 
$$(x_1\cdot 1+x_2\cdot 1)^2\le (x_1^2+x_2^2)\cdot (1^2+1^2)$$ 
or $$(x_1+x_2)^2\le 2(x_1^2+x_2^2).$$ 
This implies that
$$
\dot V\leq 3(x_1+x_2)^2 - 2(x_1^2 + x_2^2)^3\le
6(x_1^2+x_2^2)-2(x_1^2+x_2^2)^3
$$
It is easy to check that $\forall (x_1,x_2):\; x_1^2+x_2^2>\sqrt3\;\;$ $\dot V<0$, thus the solution for any initial value enters the bounded set $$
\Omega_C=\left\{  (x_1,x_2):\; x_1^2+x_2^2<C \right\}
$$
in finite time for any $C>\sqrt3$ and stays there forever, hence any solution is bounded and, due to the Bolzano–Weierstrass theorem, has a nonempty $\omega$-limit set.
</s>