= Solution
The total relativistic <energy> and <momentum> in the original frame are
$$
E=c^2(m_1\gamma_1+m_2\gamma_2),\qquad
p=m_1\gamma_1u_1+m_2\gamma_2u_2.
$$
A <Lorentz boost> of signed <velocity> $V$ makes the total spatial <momentum> $p'=\gamma_V(p-VE/c^2)$. Setting it to zero gives the <centre-of-momentum frame> <velocity>
$$
\boxed{V=\frac{c^2p}{E}
=\frac{m_1\gamma_1u_1+m_2\gamma_2u_2}
{m_1\gamma_1+m_2\gamma_2}.}
$$
It is a positive-weighted average of subluminal <velocities>, so $|V|<c$.
In the <rest mass of a coalescing relativistic collision>, <four-momentum conservation> gives $P_3=P_1+P_2$ provided no <energy> or <momentum> escapes in additional radiation. For the single outgoing particle, $u_3=c^2p/E=V$, so \b[its <speed> in the original frame is $|V|$]. Its <rest mass> is the invariant norm of the total <four-momentum>:
$$
m_3^2=\frac{E^2}{c^4}-\frac{p^2}{c^2}.
$$
Expanding and using $\gamma_i^2(1-u_i^2/c^2)=1$ gives
$$
\boxed{m_3^2=m_1^2+m_2^2
+2m_1m_2\gamma_1\gamma_2
\left(1-\frac{u_1u_2}{c^2}\right).}
$$
The positive square root is taken for $m_3$. The factor multiplying $2m_1m_2$ is the relative <Lorentz factor>, at least one. Therefore $m_3\geq m_1+m_2$, with equality when $u_1=u_2$; otherwise the initial relative <kinetic energy> becomes internal <rest energy>.
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