Problem 1112

12. Let $E(X Y Z)$ stand for the area of a triangle $X Y Z$. We have
$$\begin{array}{c} \frac{E_{1}}{E}=\frac{E(A M R)}{E(A M K)} \cdot \frac{E(A M K)}{E(A B K)} \cdot \frac{E(A B K)}{E(A B C)}=\frac{M R}{M K} \cdot \frac{A M}{A B} \cdot \frac{B K}{B C} \Rightarrow \\ \left(\frac{E_{1}}{E}\right)^{1 / 3} \leq \frac{1}{3}\left(\frac{M R}{M K}+\frac{A M}{A B}+\frac{B K}{B C}\right) . \end{array}$$

We similarly obtain
$$\left(\frac{E_{2}}{E}\right)^{1 / 3} \leq \frac{1}{3}\left(\frac{K R}{M K}+\frac{B M}{A B}+\frac{C K}{B C}\right) .$$

Therefore $\left(E_{1} / E\right)^{1 / 3}+\left(E_{2} / E\right)^{1 / 3} \leq 1$, i.e.. $\sqrt[3]{E_{1}}+\sqrt[3]{E_{2}} \leq \sqrt[3]{E}$. Analogously, $\sqrt[3]{E_{3}}+\sqrt[3]{E_{4}} \leq \sqrt[3]{E}$ and $\sqrt[3]{E_{5}}+\sqrt[3]{E_{6}} \leq \sqrt[3]{E}$; hence
$$\begin{array}{l} 8 \sqrt[8]{E_{1} E_{2} E_{3} E_{4} E_{5} E_{6}} \\ \quad=2\left(\sqrt[3]{E_{1}} \sqrt[3]{E_{2}}\right)^{1 / 2} \cdot 2\left(\sqrt[3]{E_{3}} \sqrt[3]{E_{4}}\right)^{1 / 2} \cdot 2\left(\sqrt[3]{E_{5}} \sqrt[3]{E_{6}}\right)^{1 / 2} \\ \quad \leq\left(\sqrt[3]{E_{1}}+\sqrt[3]{E_{2}}\right) \cdot\left(\sqrt[3]{E_{3}}+\sqrt[3]{E_{4}}\right) \cdot\left(\sqrt[3]{E_{5}}+\sqrt[3]{E_{6}}\right) \leq E . \end{array}$$