Why the mass of initial particle has to be greater than the sum of masses of final particles? Suppose we have a decay of a rest particle $A$ into other particles $a_1,...,a_n$
\begin{equation}
A \rightarrow a_1+a_2+\cdots+a_n
\end{equation}
It is always stated that in order to this particle decay to be possible, the mass of the initial particle $M_A$ has to be greater than the sum of masses of final particles, i.e,
\begin{equation}
M_A > M_{a_1}+\cdots+M_{a_n}=\sum_{k=1}^n M_{a_k}
\end{equation}
How can we proof this mathematically from physical principles?
I have tried using fourmomentum conservation, but I can't get to the inequality.
 A: Conservation of mass/energy tells us that
$\displaystyle M_Ac^2 + \text { kinetic energy of original particle}= \sum_{k=1}^nM_{a_k}c^2 + \text{ kinetic energy of decay products}$
If the original particle is at rest then its k.e. is zero. Assuming the decay products are not all at rest (if they were, how could we detect them ?) then the k.e. of the decay products must be greater than zero. So
$\displaystyle M_Ac^2 > \sum_{k=1}^nM_{a_k}c^2
\\ \displaystyle \Rightarrow M_A > \sum_{k=1}^nM_{a_k}$
A: At the most fundamental level [in terms of the structure of special relativity],
the reason that 
"the parent mass is greater than the sum of the daughter masses" (always the invariant masses)
is the same as

"the Clock Effect (which I will describe analogous to the above by: the proper time of the inertial path from O to Z is
greater than the sum of the proper times along a piecewise-inertial path from O to Z via an intermediate event not on OZ)".
And these are ultimately due to the "reversed triangle inequality", which involves the law of "cosines".
\begin{align}
\tilde C &= \tilde A + \tilde B\\
\tilde C\cdot \tilde C &= \tilde A\cdot \tilde A + \tilde B\cdot\tilde B + 2\tilde A\cdot \tilde B\\
C^2 &= A^2 + B^2 + 2AB \mbox{("cosine of angle between"})\\
\end{align}
[In special relativity, the time-dilation (the gamma $\gamma$) factor plays the role of [hyperbolic] cosine, as adjacent/hypotenuse, where hypotenuse is the triangle side that is opposite the "right angle" (the corner where the perpendicular sides meet).]

Next, note the algebraic fact $A^2+B^2+2AB=(A+B)^2$.


How the magnitude $C$ compares to the sum of magnitudes $(A+B)$
depends on the metric geometry,
which is essentially tied to the nature of $\mbox{("cosine of angle between"})$.
[For relativity, we restrict to future-timelike vectors.]

*

*In Euclidean space, $\mbox{("cosine of angle between"})\le 1$.
So, $C^2 = A^2 + B^2 + 2AB \mbox{("cosine of angle between"}) \le (A+B)^2$ ["triangle inequality"].

*In Minkowski spacetime, $\mbox{("cosine of angle between"})=\cosh\theta=\gamma\ge 1$.
So, $C^2 = A^2 + B^2 + 2AB \mbox{("cosine of angle between"}) \ge (A+B)^2$ ["reversed triangle inequality"... hence the Clock Effect and the Mass Effect (where the vector equation is conservation-of-total-4-momentum in energy-momentum space)].

*In Galilean spacetime, it can be argued that its $\mbox{("cosine of angle between"})=1$.
So, $C^2 = A^2 + B^2 + 2AB \mbox{("cosine of angle between"}) = (A+B)^2$ ["no triangle inequality"... hence no Clock Effect and no Mass Effect].

