= Solution
\b[Conservation laws]. Use <natural units> $c=1$ and write $p_i^\mu=(E_i,\mathbf p_i)$ with the $+---$ metric. In the parent's <rest frame>, which is the final-state <centre-of-momentum frame>, $P^\mu=(M,\mathbf0)$. <Four-momentum conservation> and the <relativistic energy-momentum relation> give
$$
\boxed{\sum_{i=1}^n E_i=M,\qquad \sum_{i=1}^n\mathbf p_i=\mathbf0,}
\qquad
E_i^2-\mathbf p_i^2=m_i^2,\quad E_i\geq m_i.
$$
A real massless daughter has strictly positive <energy>. The kinematic threshold is $M\geq\sum_i m_i$; below, no decay satisfying these conditions exists. The endpoint discussion assumes open <phase space>, $M>\sum_i m_i$, to avoid collapsed threshold configurations.
\b[Two-body energies]. <Momentum conservation> requires $\mathbf p_2=-\mathbf p_1$, so both daughters have the same momentum magnitude $p$. Their on-shell relations imply
$$
E_1^2-p^2=m_1^2,\qquad E_2^2-p^2=m_2^2,
$$
and subtraction eliminates $p$:
$$
(E_1-E_2)(E_1+E_2)=m_1^2-m_2^2.
$$
Since $E_1+E_2=M$, solving the sum and difference equations gives the <relativistic two-body decay> result
$$
\boxed{E_1=\frac{M^2+m_1^2-m_2^2}{2M},\qquad
E_2=\frac{M^2+m_2^2-m_1^2}{2M}.}
$$
As a check, their common momentum magnitude is
$$
p=\frac{\sqrt{[M^2-(m_1+m_2)^2][M^2-(m_1-m_2)^2]}}{2M}
=\frac{\sqrt{\kappa(M^2,m_1^2,m_2^2)}}{2M},
$$
where $\kappa$ is the <Källén function>. It vanishes at the massive two-body threshold and is real on the allowed side. If the symbols $M,m_i$ denote <masses> in kilograms rather than rest <energies>, multiply the displayed energy formulas by $c^2$ and the displayed momentum formula by $c$.
\b[Minimum energy in three-body decay]. The on-shell inequality already gives $E_1\geq m_1$. For $m_1>0$, the bound is attainable by putting daughter 1 at rest and letting daughters 2 and 3 recoil back to back:
$$
\mathbf p_1=0,\qquad\mathbf p_2=-\mathbf p_3=\mathbf k,\qquad
\sqrt{k^2+m_2^2}+\sqrt{k^2+m_3^2}=M-m_1.
$$
The last equation has a solution because $M-m_1>m_2+m_3$; its left side increases continuously from $m_2+m_3$ to infinity. Thus
$$
\boxed{E_1^{\min}=m_1\quad(c=1),\qquad E_1^{\min}=m_1c^2\quad\text{with units restored}.}
$$
If $m_1=0$, the same formula describes the lower boundary $E_1\to0^+$. It is an <infimum>, not an attained physical zero-energy massless particle: this is a <soft-particle endpoint>.
\b[Maximum energy from the recoil mass]. Group daughters 2 and 3 into a recoil system with <four-momentum> $Q=p_2+p_3=P-p_1$ and squared <invariant mass> $s_{23}=Q^2$. Directly in the parent <rest frame>,
$$
s_{23}=(P-p_1)^2=M^2+m_1^2-2ME_1,
\qquad
E_1=\frac{M^2+m_1^2-s_{23}}{2M}.
$$
Maximizing $E_1$ therefore means minimizing the recoil <invariant mass>. This approach does not assume a recoil rest frame when that system is null.
If both recoil daughters have positive <masses>, use their pair <rest frame>. Their total <energy> there is at least $m_2+m_3$, with equality when both are at rest relative to one another. Equivalently, their future-directed <four-velocities> satisfy $p_2\cdot p_3\geq m_2m_3$, giving
$$
s_{23}=m_2^2+m_3^2+2p_2\cdot p_3\geq(m_2+m_3)^2.
$$
Equality means the two recoil daughters have a common <velocity> in the parent frame: their momenta are parallel and $\mathbf p_2/m_2=\mathbf p_3/m_3$. Daughter 1 recoils in the opposite direction. Hence
$$
\boxed{E_1^{\max}=\frac{M^2+m_1^2-(m_2+m_3)^2}{2M}.}
$$
At this endpoint their total <energy> and its sharing are
$$
E_{23}=\frac{M^2-m_1^2+(m_2+m_3)^2}{2M},\qquad
E_2=\frac{m_2}{m_2+m_3}E_{23},\qquad
E_3=\frac{m_3}{m_2+m_3}E_{23}.
$$
These explicitly exhibit a permitted configuration attaining the bound when both recoil <masses> are positive.
If exactly one recoil daughter is massless, say $m_3=0$ and $m_2>0$, work in the <rest frame> of daughter 2. For any positive photon-like daughter energy there,
$$
s_{23}=m_2^2+2m_2E_3^{(2)}>m_2^2.
$$
The lower recoil-mass bound is approached only as $E_3^{(2)}\to0^+$. Thus the same general formula with $m_3=0$ is the \b[<supremum>] of $E_1$ for a genuine three-body event, rather than an attained maximum. Interchanging 2 and 3 covers the other mixed case. The distinction is important because the paper's category “at least one has mass” includes this <soft-particle endpoint> as well as the attainable two-massive-particle case.
If both recoil daughters are massless, instead compute
$$
s_{23}=2E_2E_3(1-\cos\theta_{23})\geq0.
$$
It can be exactly zero at nonzero $E_2,E_3$ when their momenta are parallel. Then $|\mathbf Q|=E_2+E_3$, and <momentum conservation> makes $|\mathbf p_1|=|\mathbf Q|$. Solving $E_1+|\mathbf p_1|=M$ and $E_1^2-|\mathbf p_1|^2=m_1^2$ gives
$$
\boxed{E_1^{\max}=\frac{M^2+m_1^2}{2M},\qquad
|\mathbf p_1|=E_2+E_3=\frac{M^2-m_1^2}{2M}.}
$$
Both recoil daughters can have positive <energies>: set $E_2=x|\mathbf p_1|$ and $E_3=(1-x)|\mathbf p_1|$ for any $0<x<1$. This endpoint is attained without a soft daughter, even though their combined <four-momentum> is null and has no <rest frame>.
Combining the <three-body decay energy endpoints>, the closed kinematic range is
$$
\boxed{m_1\leq E_1\leq\frac{M^2+m_1^2-(m_2+m_3)^2}{2M}\quad(c=1).}
$$
For <masses> in ordinary units, multiply both endpoint expressions by $c^2$. Open endpoints occur precisely in the soft limits described above. These bounds specify kinematic possibilities; the <decay amplitude> can still vanish at a boundary, so a possible endpoint need not have a nonzero event density.
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