= Solution
Use <natural units> and invariant squared center-of-mass <energy> $s$. For a massless incident <neutrino> and an initially stationary <electron>,
$$
s=(p_\nu+p_e)^2=m_e^2+2m_eE_\nu.
$$
At the production threshold the final massless <neutrino> can have arbitrarily small center-of-mass <energy>, so the lower endpoint is $s=m_\ell^2$. The <charged-current neutrino-electron threshold> is consequently
$$
\boxed{E_{\nu,\min}=\frac{m_\ell^2-m_e^2}{2m_e}.}
$$
For the three flavours, using precisely the masses given in this part, the answers are \b[$0$ for $\nu_e$], \b[$1.0994\times10^4\ \mathrm{MeV}\simeq11.0\ \mathrm{GeV}$ for $\nu_\mu$], and \b[$3.0898\times10^6\ \mathrm{MeV}\simeq3.09\ \mathrm{TeV}$ for $\nu_\tau$]. The zero threshold for the <electron> flavour is an infimum for elastic scattering; it does not imply a nonzero interaction rate at zero incident <energy>. The large heavy-lepton thresholds arise because most laboratory <energy> becomes center-of-mass motion against a very light target.
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