This is very easy once you compute area ratios through side length ratios. For example, we have $ [AMR]=\frac{MR}{MK}\cdot\frac{AM}{AB}\cdot\frac{BK}{BC}[ABC]$. We will find the $ \frac{CK}{BC}$ factor in the expression for $ [ALF]$. By AM-GM, $ \frac{1}{4}\ge\frac{BK}{BC}\cdot\frac{CK}{BC}$. We similarly maximize the other factors, and the answer follows.

For ease of notation, let the points $D\in LM, E\in MK, F\in KL$ We have \[\frac{[AME]}{[ABC]}=\frac{[AME]}{[AMK]}\times\frac{[AMK]}{[ABK]}\times\frac{[ABK]}{[ABC]}=\frac{ME}{MK}\times\frac{AM}{AB}\times\frac{BK}{BC}.\] Hence, by AM-GM we have \[\sqrt[3]{\frac{[AME]}{[ABC]}}\le\frac{ME/MK+AM/AB+BK/BC}{3}.\] Similarly \[\frac{[CKE]}{[ABC]}=\frac{KE}{MK}\times\frac{KC}{BC}\times\frac{MB}{AB}\] and by AM-GM \[\sqrt[3]{\frac{[CKE]}{[ABC]}}\le\frac{KE/MK+MB/AB+KC/BC}{3}.\] Adding, we get \[\sqrt[3]{[AME]}+\sqrt[3]{[CKE]}\le\sqrt[3]{ABC}.\] We repeat the same procedure for the other pairs. If we denote the areas of the six smaller triangles as $A_1, A_2, \ldots, A_6$ we have $\sum \sqrt[3]{A_i}\le 3\sqrt[3]{[ABC]}.$ Applying AM-GM again we get \[\sum\sqrt[3]{A_i}\ge 6\left(\prod\sqrt[3]{A_i}\right)^{\frac{1}{6}}.\] Hence $\sqrt[3]{[ABC]}\le 2\left(\prod\sqrt[3]{A_i}\right)^{1/6}.$ Cubing gives \[\left(\prod A_i\right)^{\frac{1}{6}}\le\frac{1}{8}[ABC].\] We are done.