Lorenz System Help Given the Lorenz system:
$$ x_1' = 2(x_2 −x_1), $$
$$x_2' = x_1(ρ −x_3)−x_2,$$
$$x_3' = x_1x_2 −5x_3 $$
with positive parameter ρ.
(c) 
By considering V = $\frac{x_1^2}{2}+x_2^2+x_3^2$ show that for ρ ∈ (0,1) the origin is asymptotically
stable.
By considering $V = ρx_1^2 +2x_2^
2 +2(x_3 −2ρ)^2$
for fixed $ρ ∈ [1,∞)$, show that all trajectories
eventually enter, and remain in, a bounded region.
(e) For fixed $ρ ∈ [1,∞)$ and $x_0 ∈ R$
Explain what you can conclude about the omega limit
set $ω(x_0)$, from (e) and the Poincare Bendixson Theorem.
My attempt
Checking that V is a lyapunov function for the equilibrium point $(0,0,0)$ on some domain $A \subset R ^3$:
1). $V(0,0,0)= 0$ - True   
2). $V(\vec{x})>0$ - True for $R^3$/{${0}$}
3). $V \in C^1$ - True for $R^3$
4). $V'<0$ -Struggling to prove true.
$V'=x_1x_1'+2x_2x_2'+2x_3x_3'=2x_1x_2(1+\rho)-O(x_i^2)$
Then $V'<0$ when $A=${$(x_1,x_2,x_3): sgn(x_1) \neq sgn(x_2)$}
 A: Writing out the full time derivative of V $$ V'=2(1+\rho)x_1x_2-2x_1^2-2x_2^2-10x_3^2.$$  Certainly, as you noticed, when $x_1$ and $x_2$ have different signs this is negative.  So then let us assume that their signs differ, so that $x_1x_2\geq0$. Since this is nonnegative we have that $$1+\rho <2 \Rightarrow 2(1+\rho)x_1x_2 < 4x_1x_2$$ So that $$V' < 4x_1x_2-2x_1^2-2x_2^2-10x_3^2=-2(x_1^2-2x_1x_2+x_2^2)-10x_3^3 = -2(x_1-x_2)^2-10x_3^2 <0.$$ 
Now this is only strictly negative when $x_1 \neq x_2$ or $x_3 \neq 0$.  So now we consider one more case: $x_1=x_2 \neq 0$ and $x_3=0$.  But then $$ V=(2(1+\rho)-4)x_1^2 = 2(\rho-1)x_1^2 < 0$$
since $\rho \in (0,1)$.  Thus $V' < 0$ in $\mathbb{R}^3 \setminus\{0\}$.
Look for a similar type calculation to show that the derivative of the second $V$ function is positive sufficiently close to the origin and negative when sufficiently far away from the origin.
I have only understood the Poincare-Bendixson Theorem to be applicable to planar systems, and so unless you have a statement of it that I have never seen, the theorem is not applicable here.  But it is worth noting that in a planar system with an attracting set that does not contain a fixed point must contain a closed orbit.  However, in higher dimensional systems, this need not be the case.  This leads one to consider strange attractors, which the Lorenz system does have.
