= Solution
For a <Kerr black hole>, the future horizon generator is $\xi=t+\Omega_H\varphi$. A future-directed causal <four-momentum> $p$ and a future-directed null vector $\xi$ satisfy $g(p,\xi)\le0$ in the $(-+++)$ convention. The conserved quantities from the <Killing vector field> are $E=-p\cdot t$ and $L=p\cdot\varphi$. Therefore
$$
\boxed{E-\Omega_HL=-p\cdot\xi\ge0.}
$$
For a massive particle the inequality is strict. Equality for a null particle requires momentum parallel to the horizon generator and describes the limiting tangent case. In a <Penrose process> the absorbed particle may have $E<0$; for $\Omega_H>0$ this requires sufficiently negative angular momentum.
In the nonextremal test-particle regime, the <Physical-process first law for a rotating black hole> identifies $\delta M=E$ and $\delta J=L$, so
$$
E-\Omega_HL=\frac{\kappa}{8\pi G}\delta A\ge0
$$
by the <second law of black-hole mechanics>. The usual hypotheses of that area law, including the <null energy condition> and the required global regularity conditions, are understood. This thermodynamic derivation is first order in the capture process; the causal <horizon flux inequality for causal matter> also applies to the extremal horizon without dividing by $\kappa$.
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