LHS
$\sin 20^{\large\circ}\sin 40^{\large\circ}\sin 80^{\large\circ}$
$\Rightarrow \large\frac{1}{2}$$(\cos 40^{\large\circ}-\cos 120^{\large\circ})\sin 20^{\large\circ}$
$\Rightarrow \large \frac{1}{4}$$(2\cos 40^{\large\circ}\sin 20^{\large\circ}-2\cos 120^{\large\circ}\sin 120^{\large\circ}$
$\Rightarrow \large\frac{1}{4}$$[\sin (40+20)^{\large\circ}-\sin (40-20)^{\large\circ}-2(-\large\frac{1}{2})$$\sin 20^{\large\circ}]$
$\Rightarrow \large\frac{1}{4}$$[\sin 60^{\large\circ}-\sin 20^{\large\circ}+\sin 20^{\large\circ}]$
$\Rightarrow \large\frac{1}{4}$$\sin 60^{\large\circ}$
$\Rightarrow \large\frac{1}{4}.\frac{\sqrt 3}{2}=\frac{\sqrt 3}{8}$
Hence proved