How to prove this inequality relating to trigonometric function? In a triangle, A, B, C are three corners of the triangle, try to prove that :
$$\root 3 \of {1 - \sin A\sin B}  + \root 3 \of {1 - \sin B\sin C}  + \root 3 \of {1 - \sin C\sin A}  \geqslant {3 \over 2}\root 3 \of 2 $$
So complicated that I have no idea...
 A: \begin{align}  
    \sin^2C &= \sin^2(A+B) = (\sin A \cos B + \cos A \sin B)^2 \\
            &= \sin^2A \cos^2B + \sin^2B \cos^2A + 2\sin A \sin B \cos A \cos B \\
            &\leqslant \sin^2A \cos^2B + \sin^2B \cos^2A + \sin A \sin B (cos^2A + cos^2B) \\
            &= \sin^2A(1-\sin^2B)+\sin^2B(1-\sin^2A)+\sin A \sin B(2-\sin^2A - \sin^2B) \\
            &= (\sin A + \sin B)^2 - 2\sin A \sin B(\sin A + \sin B)^2 \\
            &= (\sin A + \sin B)^2(1-\sin A \sin B) \\
    \Longrightarrow 1 - \sin A \sin B &\geqslant {\left( {{{\sin C} \over {\sin A + \sin B}}} \right)^2} = 2\left( {{{\sin C} \over {\sin A + \sin B}} \cdot {{\sin C} \over {\sin A + \sin B}} \cdot {1 \over 2}} \right) \\
            &= {2 \over {{{\sin A + \sin B} \over {\sin C}} \cdot {{\sin A + \sin B} \over {\sin C}} \cdot 2}} \\
            &\geqslant {2 \over {{{\left[ {{{{{\sin A + \sin B} \over {\sin C}} + {{\sin A + \sin B} \over {\sin C}} + 2} \over 3}} \right]}^3}}} \\
            &= {{27} \over 4}{\left( {{{\sin C} \over {\sin A + \sin B + \sin C}}} \right)^3} \\
    \Longrightarrow \root 3 \of {1 - \sin A\sin B}  &\geqslant {3 \over 2}\root 3 \of 2  \cdot {{\sin C} \over {\sin A + \sin B + \sin C}}  
\end{align} 
$$ \Longrightarrow \root 3 \of {1 - \sin A\sin B}  + \root 3 \of {1 - \sin B\sin C}  + \root 3 \of {1 - \sin C\sin A}  \ge {3 \over 2}\root 3 \of 2 $$
