= Solution
Use the <inverse beta decay threshold>, with the charged-lepton <mass> generalized to $m_\ell$. For a stationary <proton> and a massless incident <antineutrino>, the <Mandelstam invariant> is $s=m_p^2+2m_pE_\nu$. At threshold the final particles have no relative momentum in their <centre-of-momentum frame>, so $s_{\rm thr}=(m_n+m_\ell)^2$. Therefore
$$
E_{\nu,\rm thr}=\frac{(m_n+m_\ell)^2-m_p^2}{2m_p}=(m_n-m_p+m_\ell)+\frac{(m_n-m_p+m_\ell)^2}{2m_p}.
$$
Using the supplied, rounded <masses> gives
$$
\boxed{E_{\bar\nu_e,\rm thr}=1.813\,\mathrm{MeV},\qquad E_{\bar\nu_\mu,\rm thr}=113.1\,\mathrm{MeV}.}
$$
The second term is the recoil correction; setting the threshold equal just to the final rest-mass excess would miss it.
Reactor <electron antineutrinos> have energies of a few million <electronvolts>. They can produce a <Positron> through <inverse beta decay> above the first threshold, but cannot produce a <muon>. Thus reactor <neutrino oscillations> are measured primarily as \b[electron-antineutrino disappearance], through the event rate and energy spectrum relative to the unoscillated prediction. An oscillated <muon antineutrino> cannot be identified by the proposed muon-production channel at reactor energies; disappearance alone does not distinguish all possible final flavours.
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