= Solution
For the sextic polynomial to be bounded below at large $|m|$ while allowing either sign of $b$, its leading coefficient must obey
$$
\boxed{c>0.}
$$
If $c<0$, the <free energy> tends to $-\infty$ as $|m|\to\infty$, so it cannot define an equilibrium within this truncation. The exceptional case $c=0$ is stable for $b>0$ and gives the ordinary quartic <Landau theory>; it cannot stabilize the negative-$b$ region required for the full sextic <phase diagram>. Higher positive powers could stabilize a negative $c$, but they are absent from this model.
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