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Let $M_i$, $i=1,2,3$, be the relativistic masses of three (static) black holes and $E_i=M_i/c^2$, $i=1,2,3$ their corresponding relativistic energies. Consider the idealized situation that $E_3$ is obtained after $M_1$ and $M_2$ have merged in such a way that we can suppose the energy conservation
$$E_3=E_1+E_2.$$ Now, for the horizon area of a black hole we can write: $$A=4\pi\,r_s^2=4 \pi\left(\frac{2 G}{c^4}\right)^2\,E^2=\frac{\kappa^2}{4\pi}\,E^2,$$ while $\kappa=8\pi G/c^4$ is Einstein's gravitation constant.

From the latter expression and the energy conservation it follows that the horizon area of the final black hole is $$A_3=(A_1^{1/2}+A_2^{1/2})^2.$$ On the other hand, from the general mathematical inequality $x^{1/2}+y^{1/2}\geq (x+y)^{1/2}$, it follows that $$A_3\geq A_1+A_2.$$ That is, the area of the merged black hole is always greater than the sum of its parts before they have merged.

Is this proof reasonable?

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    $\begingroup$ When two black holes collide, the resulting outcome usually has less mass than the sum of the original two, because some of it is radiated off in gravitational waves. $\endgroup$ – Peter Shor Oct 21 '18 at 20:39
  • $\begingroup$ Related: physics.stackexchange.com/q/45448/2451 $\endgroup$ – Qmechanic Oct 23 '18 at 0:37
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This proof is perfectly fine given your assumption that the mass of the combined black hole is the sum of the masses of the two original black holes.

However, this isn't true in general ... some of the mass is usually radiated away by gravitational waves. For example, in one of the black hole collisions seen by LIGO, the two original black holes had 36 times and 29 times the mass of the Sun, and the result of the collision was a black hole with a mass of 62 times the mass of the sun. See LIGO press release.

So you haven't actually proved the area theorem for collisions of two black holes.

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  • $\begingroup$ Even considering the loss of energy by gravitational waves, always A3> A1 + A2, because this dissipated energy is much smaller than the resting energy of the smaller black hole. $\endgroup$ – João Bosco Oct 24 '18 at 4:28
  • $\begingroup$ I think that never we will get A3 = A1 + A2 $\endgroup$ – João Bosco Oct 24 '18 at 4:40
  • $\begingroup$ About this sum of areas, I see it this way. If we consider that A1 = A2 (for example) then A3 = 2 (A1 + A2), because the black hole radius is directly proportional to its mass. Then A3 = A1 + A2, only if A1 or A2 is null. Therefore A3> A1 + A2, always. $\endgroup$ – João Bosco Nov 13 '18 at 22:10
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That is, the area of the merged black hole is always greater than the sum of its parts before they have merged.

This it happens whenever the mass of an object is increased.

Note that on the Earth's surface the escape velocity is 11.2 km / s.

On the surface of the event horizon of a black hole the escape velocity is c.

If we double (for example) the mass of the Earth, the surface that still has an escape velocity of 11.2 km / s will have increased 4 times.

This result depends only the fact that the mass of an object is directly proportional to the respective radius.

EDIT;

I said respective radius, but to say radial distance is better (according with @Peter Shor)

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  • $\begingroup$ The mass of an object is directly proportional to the respective radius??? Only for black holes, not for Earth-like planets. $\endgroup$ – Peter Shor Oct 22 '18 at 19:19
  • $\begingroup$ @PeterShor The Earth's escape velocity or a black hole's escape velocity are calculated the same form, because they depends only the mass M and the distance r. Here, mass M and the distance r are directly prorportional. $\endgroup$ – João Bosco Oct 22 '18 at 23:53
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Now, for the horizon area of a black hole we can write: $$A=4\pi\,r_s^2=4 \pi\left(\frac{2 G}{c^4}\right)^2\,E^2=\frac{\kappa^2}{4\pi}\,E^2,$$ where $\kappa = 8\pi G/c^4$ is Einstein's gravitation constant.

This part is not correct. As black holes represent dramatic 'bends' in spacetime, you cannot use the spherical surface area formula; the curvature is so great there that the usual Euclidean geometry no longer holds. In fact, the 'radius' or 'mass' of a black hole usually refers to its effective radius and mass, that is, as measured by an observer far away. Blackholes do not necessarily contain any matter; they can be deduced even from vacuum, according to Einstein's equation. For instance, the Schwarzschild solution for static charge-less blackholes involves no matter.

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  • $\begingroup$ For the notion of the horizon area A see Eq. (2) in scholarpedia.org/article/… $\endgroup$ – user56224 Oct 6 '18 at 17:44

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