Solution
= Solution
Since $I'(m)=\operatorname{artanh}m$, stationarity gives
$$
\operatorname{artanh}m_e
=\beta(h+g-2Jd\,m_o),
$$
$$
\operatorname{artanh}m_o
=\beta(h-g-2Jd\,m_e).
$$
Therefore
$$
m_e=\tanh[\beta A(m_o)],
\qquad
m_o=\tanh[\beta D(m_e)],
$$
where
$$
\boxed{A(m_o)=h+g-2Jd\,m_o,
\qquad D(m_e)=h-g-2Jd\,m_e.}
$$