Solution
= Solution
For $a>0$ and $H>0$, the field tilts the single well toward positive M. Since $F''=a+3bM^2>0$ everywhere, there is exactly one minimum, at the positive solution $M_H$ of the <equation of state>. At that solution,
$$
\boxed{\langle M\rangle=M_H>0,\qquad
F_{\min}=-\frac a2M_H^2-\frac{3b}{4}M_H^4.}
$$
This follows by eliminating $\mu_0H=aM_H+bM_H^3$ from F. At weak field, $M_H\simeq\mu_0H/a$ and $F_{\min}\simeq-\mu_0^2H^2/(2a)$.