How can one prove this geometric inequality?

Let $ABCDEF$ be a convex hexagon with area $S$. Show that $$BD\cdot(AC+CE-EA)+DF\cdot(CE+EA-AC)+FB\cdot(EA+AC-CE)\ge 2\sqrt{3}\cdot S.$$

At some point, I found this similar problem.

Thank you to everyone who can help. Maybe this problem is very nice and not easy.

The problem is also stated here.

I hope to see some nice methods.

• Definitely this is not easy.. It is probably a G6.. – Apurv Jan 10 '14 at 7:22
• Reading the pages you have linked, it seems they have the solution. – Sawarnik Feb 4 '14 at 10:09

"Let $u=DZ$, $v=FX$, $t=BY$. It's not hard to see that there exists $r$ such that the circles centered at $B, D, F$ with radii $rt, ru, rv$ have a point in common. Call this point $O$. We claim $r \leq \frac {2}{\sqrt {3}}$. Since $\angle BOD + \angle DOF + \angle FOB = 360°$ one of these angles is at least $120°$, say without loss of generality $\angle FOD$. Then using the Law of Cosines on $\Delta FOD$, $\frac { - 1}{2} = \cos 120 \geq \cos \angle FOD = \frac{OF^2 + OD^2 - FD^2}{2OF\cdot OD}$ ... Now if $r > \frac {2}{\sqrt {3}}$ this rearranges to $2uv > u^2 + v^2$ contradicting the Arithmetic Mean- Geometric Mean inequality. So $r \leq \frac {2}{\sqrt {3}}$ and $O$ satisfies the conditions. Now $$AC \cdot (BD + BF - DF) = 2 AC \cdot BZ \geq 2 AC \cdot \frac{\sqrt {3}}{2} BO \geq 2\sqrt{3}[ABCO]$$ Adding up similar inequalities gives the result."