We use the fact that for any convex quadrilateral $WXYZ$, if $P$ is the intersection of the diagonals $WY$ and $XZ$, then $S_{PWX} \cdot S_{PYZ} = S_{PXY} \cdot S_{PZW}$. Let $S_{AKF} = a, S_{FLB} = f, S_{BGD}=b, S_{DHC} = d, S_{CIE} = c, S_{EJA} = e$. Notice that $S_{ALD} = \frac{(a+1)(b+1)}{f} \geqslant \frac{4\sqrt{ab}}{f}$ $S_{FGC} = \frac{(f+1)(d+1)}{b} \geqslant \frac{4\sqrt{fd}}{b}$ $S_{BHE} = \frac{(b+1)(c+1)}{d} \geqslant \frac{4\sqrt{bc}}{d}$ $S_{DIA} = \frac{(d+1)(e+1)}{c} \geqslant \frac{4\sqrt{de}}{c}$ $S_{CJF} = \frac{(c+1)(a+1)}{e} \geqslant \frac{4\sqrt{ca}}{e}$ $S_{EKB} = \frac{(e+1)(f+1)}{a} \geqslant \frac{4\sqrt{ef}}{a}$ Finally, $$3S+6 = S_{ALD}+S_{FGC}+S_{BHE}+S_{DIA}+S_{CJF}+S_{EKB} \geqslant \sum \frac{4\sqrt{ab}}{f} \geqslant 24 \implies S\geqslant 6$$

We shall show that $\triangle ABC$ and $\triangle DEF$ are the result of an affine transformation of two congruent equilateral triangles that share the same center. We use the idea of considering the centers of mass of $\triangle ABC$ and $\triangle DEF$. Let the centroids of $DEF$ and $ABC$ be $Z_1$ and $Z_2$, respectively. Let $w=\tfrac{S}{S+3}$. Note that $$Z_1 = w \cdot \text{CM}_{\text{hex}} + \frac{1-w}{3} \cdot \left(\text{CM}_{FKL}+\text{CM}_{DHG} + \text{CM}_{EJI}\right)$$where $\text{CM}_{XYZ}$ denotes the centroid of triangle $XYZ$, and similarly, $$Z_2 = w\cdot \text{CM}_{\text{hex}} + \frac{1-w}{3} \cdot \left(\text{CM}_{AKJ} + \text{CM}_{BLG} + \text{CM}_{CHI}\right)$$Thus we have that $$\frac{1}{3}\left(\overrightarrow{FA}+\overrightarrow{DB}+\overrightarrow{EC}\right) = \overrightarrow{Z_1Z_2} = \frac{(1-w)}{9}\left(\overrightarrow{FA}+\overrightarrow{LJ}+\overrightarrow{DB}+ \overrightarrow{HL}+\overrightarrow{EC} +\overrightarrow{JH}\right) = \frac{(1-w)}{9}\left(\overrightarrow{FA} + \overrightarrow{DB}+ \overrightarrow{EC}\right)$$Hence $\overrightarrow{FA}+\overrightarrow{DB} +\overrightarrow{EC} = 0$ and $Z_1=Z_2$. Now, notice that $FAJL$, $ECHJ$, and $DBLH$ are trapezoids, so for some constants $p$, $q$, $r$, we have that $0 = \overrightarrow{JL} + \overrightarrow{HJ} + \overrightarrow{LH} = p\cdot \overrightarrow{FA} + q\cdot \overrightarrow{EC} + r\cdot \overrightarrow{DB}$, which implies that $p=q=r$, and therefore $\frac{AK}{KL} = \frac{BG}{GH} = \frac{CI}{IJ}$. Similarly, we derive that $\frac{BL}{LK} = \frac{AJ}{JL} = \frac{CH}{HG}$. Now, when we take a shear transformation taking $ABC$ to be an equilateral triangle, preserving the areas of all polygons, we see that $DEF$ is also an equilateral triangle. Since the centroids of $ABC$ and $DEF$ concur, we have that, here, $DEF$ is $ABC$ rotated about their common center. From a simple calculation, it then follows that the area is minimized when $DEF$ is $ABC$ rotated by $\tfrac{\pi}{6}$. Hence $S\geqslant 6$. RemarkOne cool fact that this solution shows is that the diagonals of the hexagon concur; they concur at the common centroid.

Will fill in later.