Hope that the following solution is correct. Suppose that C>=B>=A,then sinC>= sinB>= sinA,sinAsin(A-B)sin(A-C)/sin(B+C)>=0. The inequality reduces to show that sinCsin(C-B)sin(C-A)/sin(A+B) >=sinBsin(C-B)sin(B-A)/sin(A+C). Note that sinC>= sinB,sin(C-B)>=0.So it would sufficient to show that sin(C-A)sin(C+A)>=sin(B-A)sin(B+A)is equivalent to cos2A>=cos2B<=> sin ² A>=sin ² B.

hardsoul wrote: Hope that the following solution is correct. Suppose that C>=B>=A,then sinC>= sinB>= sinA,sinAsin(A-B)sin(A-C)/sin(B+C)>=0. The inequality reduces to show that sinCsin(C-B)sin(C-A)/sin(A+B) >=sinBsin(C-B)sin(B-A)/sin(A+C). Note that sinC>= sinB,sin(C-B)>=0.So it would sufficient to show that sin(C-A)sin(C+A)>=sin(B-A)sin(B+A)is equivalent to cos2A>=cos2B<=> sin ² A>=sin ² B. Nice! Well, I didn't find any mistake except the last part, which should be $\cos{(2A)} \geq \cos{(2B)} \Leftrightarrow \sin^2{A} \leq \sin^2{B}$, which is true since $A \leq B$. Or, you could have finished the last part with by noting that $\sin{(C-A)} \geq \sin{(B-A)}$, and $\sin{(C+A)} \geq \sin{(B+A)}$. I think it's right... anyways, cool solution.

Does anyone notice that this inequality is a lot like Schur's except with trig?

as me ,I use 'Shur '.thank you problem very easy

,i too.if interested ask for sol.

Denote $a=\sin A, b=\sin B, c=\sin C$ Then it is equivalent to, $\sum \sin A\sin (A-B)\sin (A-C)\sin (A+B)\sin (A+C)\\ =\sum \sin A(\sin ^2A-\sin ^2B)(\sin ^2A-\sin ^2C)\\ =\sum a(a^2-b^2)(a^2-c^2)\\ \geq 0$ The last inequalitiy refers to Schur's.

Assume WLOG that $ A \ge B \ge C $. Note that $ \sin $ is an odd, nonnegative, monotonically increasing function in the interval $ \left(0, \frac{\pi}{2}\right) $. Moreover, since $ \frac{\sin{A}}{\sin{(B + C)}} + \frac{\sin{C}}{\sin{(A + B)}} \ge \frac{\sin{B}}{\sin{(C + A)}} $ we have that Theorem $ 3c $ in http://web.mit.edu/~darij/www/VornicuS.pdf immediately solves the problem.