= Solution
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 $A_f\geq A_1+A_2$. Define
$$
X=\frac{A_1+A_2}{8\pi}
=\sum_{i=1}^2\left(M_i^2+\sqrt{M_i^4-J_i^2}\right),\qquad
J_f=J_1+J_2.
$$
For any regular <Kerr black hole>, $|J_i|\leq M_i^2$, so $X\geq|J_1|+|J_2|\geq|J_f|$. The <irreducible mass> identity obtained by inverting the Kerr area relation is
$$
M_f^2=\frac12\left(Y+\frac{J_f^2}{Y}\right),\qquad Y=\frac{A_f}{8\pi}\geq|J_f|.
$$
Indeed, $Y=M_f^2+\sqrt{M_f^4-J_f^2}$ 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 $\tfrac12(1-J_f^2/Y^2)\geq0$ on the physical branch. Therefore the <area bound for an axisymmetric Kerr merger> is
$$
\boxed{E_{\mathrm{rad}}\leq M_1+M_2-
\sqrt{\frac12\left[X+\frac{(J_1+J_2)^2}{X}\right]}.}
$$
This 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 $(2-\sqrt2)M$, a fraction $1-1/\sqrt2$ 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.
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