if $A_{1},A_{2},A_{3},A_{4},A_{5},A_{6}$ are concyclic different points, and $A_{4}A_{6},A_{1}A_{5}$ are diameters, show that $$S_{\Delta OA_{2}A_{6}}\cdot S_{\Delta OA_{3}A_{1}}\cdot S_{\Delta OA_{4}A_{2}}\cdot S_{\Delta OA_{5}A_{3}}\ge 4S_{\Delta OA_{1}A_{2}}\cdot S_{\Delta OA_{2}A_{3}}\cdot S_{\Delta OA_{3}A_{4}}\cdot S_{\Delta OA_{4}A_{5}}$$
It seems that Ptolemy inequality could be used to solve it?Thanks
Let $\measuredangle A_0OA_1=\alpha$, $\measuredangle A_1OA_2=\beta$ and $\measuredangle A_2OA_3=\gamma$.
Hence, $\measuredangle A_3OA_4=180^{\circ}-\alpha-\beta-\gamma$ and $\measuredangle A_4OA_5=\alpha$.
Thus, we need to prove that: $$\sin^2(\alpha+\beta)\sin^2(\beta+\gamma)\geq4\sin\alpha\sin\beta\sin\gamma\sin(\alpha+\beta+\gamma)$$ and the rest is smooth.
Indeed, we need to prove that $$\left(\cos(\alpha-\gamma)-\cos(\alpha+2\beta+\gamma)\right)^2\geq8\sin\alpha\sin\gamma(\cos(\alpha+\gamma)-\cos(\alpha+2\beta+\gamma))$$ or $$x^2-2(\cos(\alpha-\gamma)-4\sin\alpha\sin\gamma)x+\cos^2(\alpha-\gamma)-8\sin\alpha\sin\gamma\cos(\alpha+\gamma)\geq0,$$ where $x=\cos(\alpha+2\beta+\gamma).$
Thus, it's enough to prove that $$(\cos(\alpha-\gamma)-4\sin\alpha\sin\gamma)^2-\left((\cos^2(\alpha-\gamma)-8\sin\alpha\sin\gamma\cos(\alpha+\gamma)\right)\leq0$$ or $$-8\cos(\alpha-\beta)\sin\alpha\sin\beta+16\sin^2\alpha\sin^2\gamma+8\sin\alpha\sin\gamma\cos(\alpha+\gamma)\leq0$$ or $$0\leq0,$$ which says that our inequality it's just $$\left(\cos(\alpha+2\beta+\gamma)-\cos(\alpha-\gamma)+4\sin\alpha\sin\gamma\right)^2\geq0.$$ Done!