amount of energy that can be extracted from the black hole by the Penrose process [@Christodoulou70]. When the difference (\[eq:4b\]) is zero, the black hole has no angular momentum and hence it is the Schwarzschild black hole.

From Newtonian considerations, we can interpret the inequality (\[eq:mj\]) as follows [@Wald71]. In a collapse the gravitational attraction ($\approx
m^2/r^2$) at the horizon ($r \approx m $) dominates over the centrifugal repulsive forces ($\approx J^2/mr^3$).

Finally, concerning the inequality (\[eq:JA\]), the black hole temperature is given by the following formula $$\label{eq:5c}
  \kappa= \frac{1}{4 m} \left(1-\frac{(8\pi J)^2 }{A^2}  \right).$$ The temperature is positive if and only if the inequality (\[eq:JA\]) holds. Moreover the temperature is zero if and only if the equality in (\[eq:JA\]) holds and hence the black hole is extreme.

There exists another relevant geometrical inequality which can be deduced from the formula (\[eq:3\]) $$\label{eq:10}
8\pi \left( m^2-\sqrt{m^4-J^2} \right) \leq  A \quad  (= \text{Extreme Kerr}).$$ Remarkably, as it was pointed out in [@Khuri:2013wha] for the case of the electric charge and in [@Dain:2013qia] for the present case of angular momentum, the inequality (\[eq:10\]) can be deduced purely from the inequalities (\[eq:mj\]) and (\[eq:JA\]) (i.e. without using the equality (\[eq:3\])) by simple algebra. 