You have one incoming slow momentum (\(P_{1}\)) and two outgoing slow momenta (\(P_{2}\) and \(P_{3}\)).
Energy-Momentum Vectors
The three external slow momenta can be written as
$$ P_{1} = \begin{pmatrix} m_{1} \cosh{(\xi_{1})} & m_{1} \sinh{(\xi_{1})} \end{pmatrix}; $$
$$ P_{2} = \begin{pmatrix} m_{2} \cosh{(\xi_{2})} & m_{2} \sinh{(\xi_{2})} \end{pmatrix}; $$
$$ P_{3} = \begin{pmatrix} m_{3} \cosh{(\xi_{3})} & m_{3} \sinh{(\xi_{3})} \end{pmatrix}. $$
Conservation
The conservation of energy leads to
$$ m_{1} \cosh{(\xi_{1})} = m_{2} \cosh{(\xi_{2})} + m_{3} \cosh{(\xi_{3})}. $$
Similarly, the conservation of momentum leads to
$$ m_{1} \sinh{(\xi_{1})} = m_{2} \sinh{(\xi_{2})} + m_{3} \sinh{(\xi_{3})}. $$
These two expressions can be combined to yield two equivalent expressions:
$$ m_{1} \exp{(\xi_{1})} = m_{2} \exp{(\xi_{2})} + m_{3} \exp{(\xi_{3})}; $$
$$ \frac{m_{1}}{\exp{(\xi_{1})}} = \frac{m_{2}}{\exp{(\xi_{2})}} + \frac{m_{3}}{\exp{(\xi_{3})}}. $$
Regge-Mandelstam Invariants
You can introduce three Regge-Mandelstam invariants:
$$ r_{12} \equiv -\frac{P_{1} \cdot P_{2}}{m_{1} m_{2}} = \frac{m_{1}^{2} + m_{2}^{2} - m_{3}^{2}}{2 m_{1} m_{2}} = \cosh{(\xi_{1} - \xi_{2})}; $$
$$ r_{13} \equiv -\frac{P_{1} \cdot P_{3}}{m_{1} m_{3}} = \frac{m_{1}^{2} - m_{2}^{2} + m_{3}^{2}}{2 m_{1} m_{3}} = \cosh{(\xi_{1} - \xi_{3})}; $$
$$ r_{23} \equiv -\frac{P_{2} \cdot P_{3}}{m_{2} m_{3}} = \frac{m_{1}^{2} - m_{2}^{2} - m_{3}^{2}}{2 m_{2} m_{3}} = \cosh{(\xi_{2} - \xi_{3})}; $$
Since the hyperbolic cosine satisfies
$$ \cosh{(x)} \geq 1; $$
you have the conditions:
$$ m_{3}^{2} \leq (m_{1} - m_{2})^{2}, \qquad m_{2}^{2} \leq (m_{1} - m_{3})^{2}, \qquad m_{1}^{2} \geq (m_{2} + m_{3})^{2}. $$
All of these are equivalent to
$$ m_{1} \geq m_{2} + m_{3}. $$
That is, the incoming mass must be larger than the sum of the two outgoing masses.
Solving for the Outgoing Data
If you introduce variables \(f_{2}\) and \(f_{3}\) via
$$ f_{2} = \exp{(\xi_{2})}, \qquad f_{3} = \exp{(\xi_{3})}; $$
then the conservation laws can be rewritten as
$$ A = m_{2} f_{2} + m_{3} f_{3}, \qquad B = \frac{m_{2}}{f_{2}} + \frac{m_{3}}{f_{3}}. $$
Here
$$ A = m_{1} \exp{(\xi_{1})}; $$
and
$$ B = \frac{m_{1}}{\exp{(\xi_{1})}}. $$
Note that
$$ A B = - \Vert P_{1} \Vert^{2} = m_{1}^{2}. $$
Solving for \(f_{2}\) and \(f_{3}\) gives
$$ f_{2} = \frac{AB + m_{2}^{2} - m_{3}^{2} + \sqrt{\left[ AB - (m_{2} + m_{3})^{2} \right] \left[ AB - (m_{2} - m_{3})^{2} \right]}}{2 m_{2} B}; $$
$$ f_{3} = \frac{AB - m_{2}^{2} + m_{3}^{2} - \sqrt{\left[ AB - (m_{2} + m_{3})^{2} \right] \left[ AB - (m_{2} - m_{3})^{2} \right]}}{2 m_{3} B}. $$
In terms of \(m_{1}\) and the variable \(f_{1}\) defined via
$$ f_{1} = \exp{(\xi_{1})}; $$
you have
$$ f_{2} = \left(\frac{m_{1}^{2} + m_{2}^{2} - m_{3}^{2} + \sqrt{\left[ m_{1}^{2} - (m_{2} + m_{3})^{2} \right] \left[ m_{1}^{2} - (m_{2} - m_{3})^{2} \right]}}{2 m_{1} m_{2}}\right) f_{1}; $$
$$ f_{3} = \left(\frac{m_{1}^{2} - m_{2}^{2} + m_{3}^{2} - \sqrt{\left[ m_{1}^{2} - (m_{2} + m_{3})^{2} \right] \left[ m_{1}^{2} - (m_{2} - m_{3})^{2} \right]}}{2 m_{1} m_{3}}\right) f_{1}. $$
Solving for the rapidities gives
$$ \xi_{2} = \xi_{1} + \log{\left(\frac{m_{1}^{2} + m_{2}^{2} - m_{3}^{2} + \sqrt{\left[ m_{1}^{2} - (m_{2} + m_{3})^{2} \right] \left[ m_{1}^{2} - (m_{2} - m_{3})^{2} \right]}}{2 m_{1} m_{2}}\right)}; $$
$$ \xi_{3} = \xi_{1} + \log{\left(\frac{m_{1}^{2} - m_{2}^{2} + m_{3}^{2} - \sqrt{\left[ m_{1}^{2} - (m_{2} + m_{3})^{2} \right] \left[ m_{1}^{2} - (m_{2} - m_{3})^{2} \right]}}{2 m_{1} m_{3}}\right)}. $$
A simpler relation between \(\xi_{2}\) and \(\xi_{3}\) follows from \(r_{23}\):
$$ \exp{(\xi_{2} - \xi_{3})} = \frac{m_{1}^{2} - m_{2}^{2} - m_{3}^{2} + \sqrt{\left[ m_{1}^{2} - (m_{2} + m_{3})^{2} \right] \left[ m_{1}^{2} - (m_{2} - m_{3})^{2} \right]}}{2 m_{2} m_{3}}. $$
