Create a trapping region for Lorenz Attractor I would like to show that the quantity: 
$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right)$
is negative on the surface:
$rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$
for some sufficiently large value of $C$. 
I was not able to massage the first quantity any more in order to make it look like the second. I also considered a change of coordinates, but had no luck.
$\sigma, b, r$  are positive parameters.
This is a step in exercise 9.2.2 from Strogatz Nonlinear Dynamics and Chaos.
 A: Since the parameter $\sigma$ is positive, the quantity 
$$
-2\sigma \left ( rx^2 + y^2 + b(z-r)^2 - br^2 \right )
$$
is negative if 
$$
rx^2 + y^2 + b(z-r)^2 > br^{2}.
$$ 
This inequality defines the exterior of an ellipsoid (call it $E_1$); note that the size of this ellipsoid is fixed by the parameters (that's a hint).  
Now the equation
$$
rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2 = C
$$
defines a different ellipsoid, $E_2$, the size of which is determined by your choice of $C$ (another hint).
At this point, remind yourself what it is that you want to show.  Typically, the goal is to show that there exists a $C$ such that $E_2$ defines a trapping region for the Lorenz equations, in which case it suffices to show that  $E_2$ can be made large enough so that it contains $E_1$.  There's really no additional calculation necessary to do this -- you just need to understand what you've done so far.  
A related (but different) question is to find an explicit lower bound on $C$ in terms of the parameters.  In this case, you can find bounds on each of $x$, $y$, and $z$ separately for points inside of $E_1$.  This will then give you a bound on the quantity
$$
rx^2 + \sigma y^2 + \sigma \left ( z - 2r \right )^2
$$
which then defines $C$.
A: Just simplifying:
$-2\sigma\left(rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2}\right) < 0$
$=> rx^{2}+y^{2}+b\left(z-r\right)^{2}-br^{2} > 0 ... \sigma > 0$
$=> rx^{2}+y^{2}+bz^{2}-2brz+br^{2}-br^{2} > 0 $
$=> rx^{2}+y^{2}+bz^{2}-2brz > 0 $
$=> rx^{2}+y^{2}+bz^{2} > 2brz $
$=> z<=0$ or $z>=2r$ for any $C$ or
$ rx^{2}+y^{2} > 2brz - bz^{2}, 0<z<2r $
Now,
$rx^{2}+\sigma y^{2}+\sigma\left(z-2r\right)^{2}=C$
$=> rx^{2}+y^{2}+(\sigma-1) y^{2}+\sigma\left(z-2r\right)^{2}=C$
$=> rx^{2}+y^{2}= C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}$
So the original expression is negative when $z$ is outside $(0,2r)$ or
$ 2brz - bz^{2} < C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0<z<2r $
$=> bz(2r - z) < C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}, 0<z<2r $
$=> 0 < C -(\sigma-1) y^{2}-\sigma\left(z-2r\right)^{2}+bz(z - 2r), 0<z<2r $
$=> 0 < C -(\sigma-1) y^{2}-\sigma\left(z(1-b/2)-2r\right)^{2}+(bz/2)^{2}, 0<z<2r $ ... completing the square
