special relativity - Mathematical proof that the centre of mass energy is the smallest energy in a collision

I recently answered this question about how the energy of a system of colliding particles is minimised in the centre of mass frame.

I'm happy with the qualitative answer that treating the collection of particles as a whole and considering a "ball" of momentum equal to the total momentum, then since the mass is invariant, the lowest energy is achieved when this ball is not moving (i.e. in the centre of mass frame).

However, in my answer I attempted to show this explicitly for a two-particle head-on collision using the energy $$E=\sqrt{m_1^2+|\vec{p}_1|^2}+\sqrt{m_2^2+|\vec{p}_2|^2}$$and quickly found that the algebra required to get any meaningful results got messy (note that I'm working in units with $c=1$).

My question is is there a nice "quick" way to show that the energy is indeed minimised in the centre of mass frame? Am I missing some important trick that may be useful? If not, do I simply have to slog through the algebra and eventually get the answer?

Answer

The four-momentum norm is invariant \begin{align} E_{A}^{2} - p_{A}^{2} \; = \; E_{B}^{2} - p_{B}^{2} \end{align} where $E$ is energy and $p$ is the magnitude of three-momentum. The subscript $A$ and $B$ represents different inertial frame. Here $c = 1$.

Now, we choose $A$ to be the centre of mass frame (COM) such that $p_{A} = p_{COM} = 0$. Hence, we have \begin{align} E_{COM}^{2} \; &= \; E_{B}^{2}-p_{B}^{2} \\ E_{COM}^{2} - E_{B}^{2} \; &= \;-p_{B}^{2} \; < \; 0\\ E_{COM}^{2} \; &< \; E_{B}^{2} \\ E_{COM} \; &< \; E_{B} \\ \end{align} for all inertial frame $B$ with $p_{B} \neq 0$