 Solution

For the sextic polynomial to be bounded below at large $|m|$ while allowing either sign of $b$, its leading coefficient must obeyIf $c<0$, 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.