In acute $\Delta ABC$ if$$\sin{A}=2\sin{B}\sin{C}$$ show that$$\tan{A}\tan{B}\tan{C}\ge 8$$
Here is a solution
since $$\sin{A}=\sin{(B+C)}=\sin{B}\cos{C}+\sin{C}\cos{B}=2\sin{B}\sin{C}$$$$\Longrightarrow \tan{B}+\tan{C}=2\tan{B}\tan{C}$$ so $$x=\tan{A}\tan{B}\tan{C}=\tan{A}+\tan{B}+\tan{C}\ge \tan{A}+2\tan{B}\tan{C}\ge 2\sqrt{2x} $$$$\Longrightarrow x\ge 8$$
Is there any other method? Especially I would like to see a geometric solution.