If $\sigma, \tau, \beta >0$ then all solutions of the Lorenz equation \begin{eqnarray*} \dot{x} & = & \sigma(y-x)\\ \dot{y} & = & x(\tau - z) -y\\ \dot{z} & = & xy - \beta z \end{eqnarray*}will enter an ellipsoid centered at $(0,0,2\tau )$ in finite time. In addition the solution will remain inside the ellipsoid once it has entered. To observe this we define a Lyapunov function $$V(x,y,z)=\tau x^2 + \sigma y^2 + \sigma (z-2\tau )^2.$$ It then follows that \begin{eqnarray*} \dot{V} & = & 2\tau x\dot{x} + 2\sigma y\dot{y} + 2\sigma (z-2\tau )\dot{z}\\ & = & 2\tau x\sigma(y-x) + 2\sigma y(x(\tau - z) -y) + 2\sigma (z-2\tau )(xy - \beta z)\\ & = & -2\sigma (\tau x^2 + y^2 + \beta(z -r)^2 -b\tau^2). \end{eqnarray*}We then choose an ellipsoid which all the solutions will enter and remain inside. This is done by choosing a constant $C>0$ such that the ellipsoid $$\tau x^2 + y^2 + \beta(z -r)^2 = b\tau^2$$ is strictly contained in the ellipsoid $$\tau x^2 + \sigma y^2 + \sigma (z-2\tau )^2=C.$$ Therefore all solution will eventually enter and remain inside the above ellipsoid since $\dot{V}<0$ when a solution is located at the exterior of the ellipsoid.