= Solution
For a spherical homogeneous <nucleus>, the capillarity approximation gives
$$
\Delta G(r)=4\pi r^2\gamma+\frac{4\pi r^3}{3}\Delta g_V.
$$
The first term penalizes creating an interface; below the melting point the bulk free-energy change per <volume> $\Delta g_V$ is negative and favors growth. Near $T_m$, neglect the <temperature> dependence of the <latent heat>. Since the free-energy difference vanishes at $T_m$,
$$
\Delta g_V\simeq\Delta H_V\frac{\Delta T}{T_m},
\qquad\Delta T=T_m-T>0,
\qquad\Delta H_V<0\quad\text{on freezing}.
$$
Set the <radius> derivative to zero:
$$
\frac{\mathrm d\Delta G}{\mathrm dr}=8\pi r\gamma+4\pi r^2\Delta g_V=0,
\qquad
\boxed{r^*=-\frac{2\gamma}{\Delta g_V}
=-\frac{2\gamma T_m}{\Delta H_V\Delta T}.}
$$
The nonzero stationary point is a maximum, since $\Delta G''(r^*)=-8\pi\gamma<0$. Subcritical <nuclei> lower free energy by shrinking; supercritical <nuclei> lower it by growing. The positive <radius> and its inverse-supercooling dependence follow from the negative freezing <enthalpy>. This derivation assumes an isotropic, size-independent interfacial energy and a bulk-like spherical <nucleus>; it is not a molecular-scale kinetic theory.
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