Convex cyclic hexagon $ABCDEF$. Prove $AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA$

Question

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.

Answer 1

Nice problem!

Let me post a solution using inversion and cross-ratios.

One can rewrite the inequality in the following way: $$(ABCF)\cdot(BCDA)\cdot(CDEB)\cdot(DEFC)\cdot(EFAD)\cdot(FABE)\ge 729 \qquad (\heartsuit)$$ where for brevity we write $(XYZT)$ for the crossratio $(X,Y;Z,T)$.

Consider an inversion with respect to a circle centered at $F$. Let the images of $A,B,C,D,E$ be $A', B', C', D', E'$, respectively. By basic properties of inversion these points lie on a common line, say $\ell$. Denote the point at infinity of $\ell$ by $F'$. For every quadruple $X,Y,Z,T$ such that $(XYZT)$ appears in $(\heartsuit)$ we have $(XYZT)=(FX,FY;FZ,FT)=(FX',FY';FZ',FT')=(X'Y'Z'T')$ where $FF$ is understood as the line tangent to the circumcircle of $ABCDEF$ at $F$. Therefore we have to prove a variant of $(\heartsuit)$ in which every letter $X$ is replaced by $X'$; call the new inequality $(\spadesuit)$.

Since $ABCDEF$ is convex, points $A',B',C',D',E'$ lie on $\ell$ in this order. Denote $2x=A'B', y=B'C', z=C'D', 2t=D'E'$. Then $(\spadesuit)$ can be written as $$\frac{(z+2t)(y+z)(2x+y)(2x+y+z+2t)}{xyzt}\ge 108.$$ This follows from AM-GM: just multiply the following: \begin{align*} z+2t &\ge 3z^{1/3}t^{2/3}, \\ y+z &\ge 2y^{1/2}z^{1/2}, \\ 2x+y &\ge 3x^{2/3}y^{1/3}, \\ 2x+y+z+2t &\ge 6x^{2/6}y^{1/6}z^{1/6}t^{2/6}. \end{align*}

Answer 2

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.