Further explanation of the Penrose Conjecture

I'm currently a third year maths undergrad, writing a dissertation on the application of minimal surfaces in space.

I have recently come across the Penrose Conjecture that the mass of a spacetime is:

$$m \geq \sqrt{\dfrac{A}{16 \pi}}.$$

What significance does this have in terms of the geometry of a black hole? Also how does this directly relate to that of minimal surfaces? In that, is it purely to do with the apparent horizons corresponding to a minimal surface?

• See arXiv:math/9911173 [math.DG]. Though I have not managed to find time to begin reading the paper, the abstract's second paragraph talks quite a lot about Schwarzschild black holes. – user28355 Feb 2 '14 at 16:25
• Sure, no problem! – user28355 Feb 2 '14 at 16:28
• A Schwarschild black hole of mass $m$ will have an event horizon of radius $2m$ (in units where $\hbar = c = 1$). So the surface area of the black hole is $A = 4 \pi r^2 = 16 \pi m^2$. So any mass $m$ will "gobble up" at least "that much" space. That's the relation to black holes. – Siva Feb 2 '14 at 16:57