= Solution
The physical argument for the <Penrose inequality> combines <weak cosmic censorship conjecture>, the <dominant energy condition>, and relaxation to a stationary <black hole>. Work in <geometrized units>. Let $M_f$ and $A_f$ be the final <Kerr black hole> mass and horizon area. Positive energy radiated to infinity gives $E_{\rm ADM}\geq M_f$, where $E_{\rm ADM}$ is the initial <ADM energy>. For a <Kerr black hole> with $a_f=J_f/M_f$,
$$
A_f=8\pi M_f\left(M_f+\sqrt{M_f^2-a_f^2}\right)\leq16\pi M_f^2.
$$
If the initial <apparent horizon> obeys the necessary <apparent-horizon area comparison> with the enclosing <event horizon>, and <Hawking's area theorem> applies during the evolution, then
$$
A_{\rm app}\leq A_{\rm EH,initial}\leq A_f\leq16\pi M_f^2\leq16\pi E_{\rm ADM}^2.
$$
Consequently the anticipated answer, under those additional hypotheses, is
$$
\boxed{E_{\rm ADM}\geq\sqrt{\frac{A_{\rm app}}{16\pi}}.}
$$
The bound is saturated by a nonrotating <Schwarzschild black hole> with no energy loss. Rotation or outgoing radiation makes the argument's inequalities stricter.
There is an essential qualification: inclusion inside an <event horizon> does not by itself compare areas. An arbitrary <apparent horizon> on general, non-time-symmetric initial data need not satisfy the displayed <apparent-horizon area comparison>; the unqualified version with its area is not universally true, even with the <dominant energy condition>. On time-symmetric data the relevant outermost <minimal surface> is an <outer area-minimizing surface>, as used in the <Riemannian Penrose inequality>, with nonnegative <scalar curvature>. In more general formulations an appropriate enclosing-area quantity is needed. \b[The physical expectation is conditional on this area comparison], as well as on censorship, predictability, settling, and the energy assumptions; the mere presence of a <trapped surface> does not supply every step.
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