= Solution
Use <metric signature> $(+---)$, as required by the sign of the scalar product in this question. For a massive particle, the <four-momentum> is
$$
\boxed{P=(E/c,p)=m\gamma(c,v),\qquad\gamma=(1-|v|^2/c^2)^{-1/2}.}
$$
For a <photon> of positive frequency $\nu$ and direction $e$, the <Planck constant> gives $E=h\nu$ and
$$
\boxed{K=\left(\frac{h\nu}{c},\frac{h\nu}{c}e\right),\qquad K^2=0.}
$$
A <future-pointing timelike four-vector> $P=(P^0,p)$ has $P^0>|p|$ and $P^0>0$ in this signature. The ordinary Euclidean <Cauchy-Schwarz inequality> yields
$$
P_1\cdot P_2=P_1^0P_2^0-p_1\cdot p_2\geq P_1^0P_2^0-|p_1||p_2|>0.
$$
This is stronger than the requested nonnegative inequality. For an electron and a positron of rest <mass> $m>0$, their future-pointing timelike <four-momenta> have $P_1^2=P_2^2=m^2c^2$. Thus
$$
(P_1+P_2)^2=2m^2c^2+2P_1\cdot P_2>0,
$$
which cannot equal the null norm of a single <photon>. \b[<Four-momentum conservation> forbids spontaneous <photon> decay into an electron-positron pair in vacuum.] An external body or field able to exchange momentum would be a different system.
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