 Solution

The proposed Landau free energy has stationary equation $m[a(T-T_c)+b{m}^2]=0$. For $T>T_c$ its minimum is $m=0$. For $T<T_c$ the minima are $\boxed{m=\pm \sqrt{a(T_c-T)/b}}$, while zero becomes a maximum; at $T_c$ the positive quartic term keeps the free energy bounded below. Thus the two symmetry-related states, continuous transition and square-root onset agree with the mean-field approximation.

More explicitly the mean-field entropy expansion gives $F=F_0+N[k_B(T-T_c)m^2/2+k_{B}T{m}^4/12+O(m^6)]$ at zero field. Near $T_c$, replacing $T$ by $T_c$ in the quartic coefficient gives the positive constants $a=N{k}_B$ and $b=N{k}_B{T}_c/3$, and reproduces the preceding square-root amplitude. The polynomial is a local expansion near the transition, not a global expression at every temperature.