= Solution
The <four-momentum> is
$$
P=(P^0,\mathbf P)=(\gamma mc,\gamma m\mathbf u),
\qquad
\gamma=(1-|\mathbf u|^2/c^2)^{-1/2}.
$$
Thus
$$
P^0=mc+\frac{m|\mathbf u|^2}{2c}+\frac{3m|\mathbf u|^4}{8c^3}+O(|\mathbf u|^6/c^5).
$$
Multiplication by $c$ identifies the first two terms as the <rest energy> $mc^2$ and the Newtonian <kinetic energy> $m|\mathbf u|^2/2$.
For the fixed-target reaction, the initial invariant is
$$
(P_1+P_2)^2=2m^2c^2(1+\gamma).
$$
At threshold the four final particles are at rest in their centre-of-momentum frame, giving invariant $(4mc)^2$. Hence $\gamma=7$ and the least laboratory speed is
$$
\boxed{u=c\sqrt{1-\frac1{49}}=\frac{4\sqrt3}{7}c.}
$$
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