Convex cyclic hexagon $ABCDEF$. Prove $AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA$ 
Convex hexagon $ABCDEF$ inscribed within a circle. Prove that
$$AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA\,.$$

I was thinking of represending the inequalities in trigonometry then use Language multiplier. For example let $\angle AOB = \theta_1$, $\angle BOC = \theta_2$, represent the inequality in trigonometry, subject to constraint $\theta_1 + \theta_2 + ... + \theta_6 = 2\pi$. But it's still quite a bit of work and I didn't manage to get to the end. It also seems a bit overkill -- might be better solution? Would like to see any approach.

This problem had been posted here, but was deleted by the owner for an unspecified reason.  I flagged the moderators about this, but they didn't do anything.  Here is the same solution I gave in that link.
Ptolemy's theorem with $\square ABCD$ yields
$$AB\cdot CD+AD\cdot BC=AC\cdot BD.$$
Ptolemy's theorem with $\square ACDE$ yields
$$AC\cdot DE+EA\cdot CD=AD\cdot CE.$$
Hence,
$$AD=\frac{AC\cdot DE+EA\cdot CD}{CE}$$
so that
$$AC\cdot BD=AB\cdot CD+AD\cdot BC=AB\cdot CD+\left(\frac{AC\cdot DE+EA\cdot CD}{CE}\right)\cdot BC.$$
Hence
$$AC\cdot BD=AB\cdot CD+\frac{AC}{CE}(BC\cdot DE)+\frac{EA}{CE}(BC\cdot CD).$$
By AM-GM,
$$AC\cdot BD\geq 3\sqrt[3]{(AB\cdot CD)\left(\frac{AC}{CE}(BC\cdot DE)\right)\left(\frac{EA}{CE}(BC\cdot CD)\right)}=3\sqrt[3]{AB\cdot BC^2\cdot  CD^2\cdot DE\cdot \frac{AC\cdot EA}{CE^2}}.$$
This shows that
$$\sqrt[3]{\frac{AC^2\cdot BD^3\cdot CE^2}{EA}}\geq 3\sqrt[3]{AB\cdot BC^2\cdot CD^2\cdot DE}.$$
Similarly,
$$\sqrt[3]{\frac{BD^2\cdot CE^3\cdot DF^2}{FB}}\geq 3\sqrt[3]{BC\cdot CD^2\cdot DE^2\cdot EF},$$
$$\sqrt[3]{\frac{CE^2\cdot DF^3\cdot EA^2}{AC}}\geq 3\sqrt[3]{CD\cdot DE^2\cdot EF^2\cdot FA},$$
$$\sqrt[3]{\frac{DF^2\cdot EA^3\cdot FB^2}{BD}}\geq 3\sqrt[3]{DE\cdot EF^2\cdot FA^2\cdot AB},$$
$$\sqrt[3]{\frac{EA^2\cdot FB^3\cdot AC^2}{CE}}\geq 3\sqrt[3]{EF\cdot FA^2\cdot AB^2\cdot BC},$$
and
$$\sqrt[3]{\frac{FB^2\cdot AC^3\cdot BD^2}{DF}}\geq 3\sqrt[3]{FA\cdot AB^2\cdot BC^2\cdot CD}.$$
Multiplying all six inequalities above gives
$$(AC\cdot BD\cdot CE\cdot DF\cdot EA\cdot FB)^2\geq (27\cdot AB\cdot BC\cdot DE\cdot EF\cdot FA)^2\,,$$
which is equivalent to the required inequality.  The equality holds if and only if $ABCDEF$ is a regular hexagon.
