= Solution
Average over the two initial electron spins and two initial photon polarizations, and sum over the final ones. The supplied <fermion spin sum> and photon polarization sum turn the result into <gamma-matrix traces>. The <Clifford algebra> implies
$$
\gamma_\mu\not a\gamma^\mu=-2\not a,
\qquad
\gamma_\mu\not a\not b\gamma^\mu=4a\cdot b,
$$
by anticommuting the outside matrix through the product. Together with
$$
\operatorname{tr}(\not a\not b\not c\not d)
=4(a\cdot b\,c\cdot d-a\cdot c\,b\cdot d+a\cdot d\,b\cdot c),
$$
the two channel squares reduce, at $m=0$, to $-2e^4u/s$ and $-2e^4s/u$; the remaining cross terms cancel. Thus, in terms of the <Mandelstam variables>,
$$
\boxed{\overline{|\mathcal M|^2}
=-2e^4\left(\frac{s}{u}+\frac{u}{s}\right).}
$$
Physical Compton kinematics has $s>0$ and $u<0$, so the displayed expression is nonnegative.
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