these solutions are called extreme black holes. In the white region, excluding the horizon axis, the Kerr solution contains a naked singularity. That includes also the negative mass region. []{data-label="fig:2"}](kerr-parametros-p.pdf){width="70.00000%"}

The area of the horizon of the Kerr black hole is given by the simple but very important formula $$\label{eq:3}
  A=8\pi \left(m^2+ \sqrt{m^4-J^2}  \right).$$ From equation (\[eq:3\]) we deduce that the following three geometric inequalities hold for a Kerr black hole $$\begin{aligned}
 \sqrt{\frac{A}{16\pi}} &\leq m &(=\text{Schwarzschild}),\label{eq:pen}\\
   \sqrt{|J|} &\leq  m  &(= \text{Extreme Kerr}),\label{eq:mj}\\
    8\pi |J| &\leq  A  &(= \text{Extreme Kerr}).\label{eq:JA}\end{aligned}$$ As expected from the discussion above, the inequality (\[eq:mj\]) is needed to define the black hole horizon area in (\[eq:3\]): if (\[eq:mj\]) does not hold, then the expression (\[eq:3\]) is not a real number. We have listed this inequality again here to emphasize its connection with the other two in the following discussion. Inequalities (\[eq:pen\]) and (\[eq:JA\]) follow from (\[eq:mj\]) and (\[eq:3\]). Note that these inequalities relate the three relevant parameters of the Kerr black hole $(m,J,A)$.

Let us discuss the physical meaning of the inequalities (\[eq:pen\]), (\[eq:mj\]) and (\[eq:JA\]). In the inequality (\[eq:pen\]), the difference $$\label{eq:4b}
 m-\sqrt{\frac{A}{16\pi}},$$ represents the rotational energy of the Kerr black hole. This is the maximum 