# 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...

• @Macavity Yes, I have made some changes on the question. – Vlander CC Oct 19 '14 at 9:51
• A similar question. – Lucian Oct 19 '14 at 10:15

\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} $$\Longrightarrow \root 3 \of {1 - \sin A\sin B} + \root 3 \of {1 - \sin B\sin C} + \root 3 \of {1 - \sin C\sin A} \ge {3 \over 2}\root 3 \of 2$$