holds, then by the discussion above we have that the quasi-local mass ${m_{bh}}$ increase with time, that is $$\label{eq:8b}
  {m_{bh}}\leq m_0.$$ We emphasize that this inequality is highly non-trivial. The quantity ${m_{bh}}$ is computed on the initial surface $S$, in contrast to compute $m_0$ we need to known the whole spacetime. Using (\[eq:8b\]) and (\[eq:4\]) we finally obtain $$\label{eq:6}
\sqrt{\frac{A}{16\pi}+\frac{4\pi J^2}{A}}=  {m_{bh}}\leq m.$$ This inequality has the natural interpretation that the mass of the black hole ${m_{bh}}$ should always be smaller than the total mass of the spacetime $m$. The inequality (\[eq:6\]) represents a generalization of the Penrose inequality with angular momentum. This inequality implies $$\label{eq:mjd}
  \sqrt{|J|}\leq m.$$ In fact, the inequality can be deduced directly by the same heuristic argument without using the area theorem. It depends only on the following assumptions

-   Gravitational waves carry positive energy.

-   Angular momentum is conserved in axial symmetry.

-   In a gravitational collapse the spacetime settles down to a final Kerr black hole.

Let us summarize the discussion of this section. For an axially symmetric, dynamical black hole, the following two geometrical inequalities are expected $$\begin{aligned}
    8\pi |J| & \leq A \quad (=\text{Extreme Kerr horizon}), \label{eq:JAd}\\
    \sqrt{\frac{A}{16\pi}+\frac{4\pi J^2}{A}} & \leq m 