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For initially well-separated Kerr black holes with no orbital angular momentum, exact axisymmetry gives $J_f=J_1+J_2$. Put $X=(A_1+A_2)/(8\pi )$. The black-hole area theorem and the irreducible mass identity implyIndeed, $X\ge |J_1|+|J_2|\ge |J_f|$, and $\frac{1}{2}(Y+J_f^2/Y)$ is nondecreasing for $Y\ge X$. Opposite spin signs lower $J_f^2$ at fixed initial areas and hence raise the allowed radiated-energy bound; this need not be achievable in an actual merger.

The irreducible mass of a four-dimensional uncharged Kerr black hole is $M_{\mathrm{irr}}=\sqrt{A/(16\pi )}$. Its mass satisfies $M^2=M_{\mathrm{irr}}^2+J^2/(4M_{\mathrm{irr}}^2)$. Hawking's area theorem prevents a classical process from decreasing $M_{\mathrm{irr}}$ under the theorem's energy and global hypotheses.

Assume the classical black-hole area theorem applies: the null energy condition and the required global regularity or strong asymptotic predictability hypotheses hold. The disconnected initial horizon sections then give $A_f\ge A_1+A_2$. DefineFor any regular Kerr black hole, $|J_i|\le M_i^2$, so $X\ge |J_1|+|J_2|\ge |J_f|$. The irreducible mass identity obtained by inverting the Kerr area relation isIndeed, $Y=M_f^2+\sqrt{M_f^4-J_f^2}$ implies $(Y-M_f^2{)}^2=M_f^4-J_f^2$, which gives this expression. At fixed $J_f$, its derivative with respect to $Y$ is $\frac{1}{2}(1-J_f^2/Y^2)\ge 0$ on the physical branch. Therefore the area bound for an axisymmetric Kerr merger isThis uses the ideal limit of infinite initial separation, in which the initial total energy is $M_1+M_2$. At finite separation it is the initial total gravitational energy, including binding energy, that replaces this sum. The inequality is an upper limit, not a prediction or a guarantee of saturation; real mergers increase the horizon area.

Changing only a spin sign leaves $X$ unchanged, whereas the square of the final angular momentum changes. For fixed nonzero $|J_1|,|J_2|$, same-sign spins give $(|J_1|+|J_2|{)}^2$, and opposite signs give $(|J_1|-|J_2|{)}^2$. The remnant must retain more rotational energy in the same-sign case, so its minimum allowed mass is larger and the radiated-energy upper limit is smaller. Cancellation of opposite spins permits more of the initial rotational energy to be radiated while respecting the same initial total area.

For example, equal extremal initial holes have $M_1=M_2=M$ and $|J_1|=|J_2|=M^2$. Same-sign spins allow at most $(2-\sqrt{2})M$, a fraction $1-1/\sqrt{2}$ of the initial energy; opposite signs allow at most $M$, a fraction $1/2$. These are geometric bounds under exact axisymmetry, not achievable-efficiency claims.