Trigonometry Inequality

[Tassandro]

Proof that [MATH]cos\left(\frac{B-C}{2}\right)cos\left(\frac{C-A}{2}\right)cos\left(\frac{A-B}{2}\right) \geq senAsenBsenC[/MATH].[MATH]⁷[/MATH], for A,B,C,angles of a sharp triangule.

 

[Tassandro]

First, I used the identity [MATH]\sin A\sin B\sin C=\frac{\sin2A+\sin2B+\sin2C}{4}[/MATH]. Then, I did [MATH]\cos(\frac{B-C}{2})\cos(\frac{C-A}{2})=\cos(\frac{B-A}{2})+\cos(B+A-2C)[/MATH][/sub]

 

[Tassandro]

So, I used [MATH]\cos (\frac{B-A}{2})+\cos(B+A-2C)=\cos(\frac{A-B}{2})+\cos(\frac{π-3C}{2})[/MATH],but I don't know what to do right now.