= Solution
In the <pion> <rest frame>, write its <four-momentum> as $P=(m_\pi,\mathbf0)$ and the charged lepton's as $p_\ell=(E_\ell,\mathbf p)$. By <four-momentum conservation>, the massless <antineutrino> has $k=P-p_\ell$, with $k^2=0$. Therefore
$$
0=(P-p_\ell)^2=m_\pi^2+m_\ell^2-2m_\pi E_\ell,
\qquad E_\ell=\frac{m_\pi^2+m_\ell^2}{2m_\pi}.
$$
Its energy is $E_\nu=m_\pi-E_\ell$, and for a massless <antineutrino>, $E_\nu=|\mathbf k|$. The final <momenta> are equal and opposite, so
$$
\boxed{|\mathbf p_\ell|=|\mathbf k|=E_\nu
=\frac{m_\pi^2-m_\ell^2}{2m_\pi}.}
$$
The expression is positive precisely when the two-body decay is energetically allowed, $m_\pi>m_\ell$, neglecting the <neutrino mass>.
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