### 2020 International Zhautykov Olympiad #3

Given convex hexagon $ABCDEF$, inscribed in the circle. Prove that $$AC*BD*DE*CE*EA*FB \geq 27 AB * BC * CD * DE * EF * FA$$

Let $$X = AC\cdot BD\cdot CE\cdot DF\cdot EA\cdot FB \quad \text{and}\quad Y=AB \cdot BC\cdot CD\cdot DE \cdot EF\cdot FA$$By Ptolemy's theorem in quadrilateral $ABCD$ we obtain $$AC \cdot BD = AB\cdot CD + BC \cdot AD = AB\cdot CD + \frac{BC \cdot AD}{2} + \frac{BC \cdot AD}{2}$$ $$\geq 3\sqrt[3]{\frac{AB \cdot BC^2\cdot CD\cdot AD^2}{4}}$$We can get similar inequalities by applying Ptolemy's theorem in quadrilaterals in $BCDE$, $CDEF$, $DEFA$, $EFAB$, $FABC$. Multiplying these six inequalities side by side we get $$X \geq \frac{27}{4}\cdot \left(Y\cdot AD \cdot BE \cdot CF\right)^{2/3}$$Then, using Ptolemy's theorem for quadrilateral $ABCE$ we get $$AC \cdot BE = AB\cdot CE + BC \cdot EA \geq 2\sqrt{AB \cdot BC\cdot CE\cdot EA}$$Writing the same inequality for quadrilaterals $BCDF$, $CDEA$, $DEFB$, $EFAC$, $FABD$ and multiplying the resulting inequalities side by side we obtain $$AD \cdot BE \cdot CF \geq 8 \cdot Y^{1/2}$$So $$X \geq \frac{27}{4}\cdot \left(Y\cdot AD \cdot BE \cdot CF\right)^{2/3} \geq \frac{27}{4}\cdot \left(Y\cdot 8 \cdot Y^{1/2}\right)^{2/3} = 27\cdot Y,$$ as desired. $\square$