Let $E'$ and $F'$ be the points on $AB$ and $AC$ such that $E'$, $G$, $F'$ are collinear and $E'F' \parallel EF$. Since $\triangle EFG$ and $\triangle EFE'$ have the same base $EF$ and the same height, they have the same area. This yields
$$[BEGF] = [BEF] + [EFG] = [BEF] + [EFE'] = [BFE'] \ge [BF'E']$$

The last inequality holds since $\triangle BFE'$ and $\triangle BF'E'$ have the same base $BE$, but $F'$ is closer to the sideline $BE'$ than $F$. Equality holds when $F = F'$, i.e. $D = G$. Similarly, we have $[CFGE] \ge [CE'F']$, with equality $D = G$. Thus, we have
$$[BEGF] + [CFGE] \ge [BF'E'] + [CE'F'] = 2[MF'E'] = [AE'F']$$
since $M$ is the midpoint of $BC$ and $AG : GM = 2 : 1$. It suffices to prove $[AE'F'] \ge \frac{4}{9}[ABC]$.

Let $B'$ and $C'$ be the points on $AB$ and $AC$ such that $B', G, C'$ are collinear and $B'C' \parallel BC$. As $\triangle AB'C' \sim \triangle ABC$ with ratio $2:3$, we have $[AB'C'] = \frac{4}{9}[ABC]$.
Therefore, it remains to show $[AE'F'] \ge [AB'C']$. WLOG assume $E'$ lies on the segment $BB'$ and $F'$ lies on the segment $AC'$. Then we only need to show $[GB'E'] \ge [GC'F']$. Indeed, let $P$ be the symmetric point of $F'$ in $G$. Then $\triangle B'GP \cong \triangle C'GF'$, so that $P$ lies on the line $F'G$. Also, $P$ must lie on the segment $GE'$ since
$$\angle GPB' = \angle GF'C' = \angle F'E'A + \angle E'AF' > \angle PE'B'$$
It follows that $[GB'E'] \ge [GB'P] = [GC'F']$ as desired. Equality holds when $E = B'$ and $F = C'$, i.e. $D = G$ and $EF \parallel BC$.