TITLE: How prove this $\frac{\sin{(A-B)}\sin{(A-C)}}{\sin{2A}}+\frac{\sin{(B-C)}\sin{(B-A)}}{\sin{2B}}+\frac{\sin{(C-A)}\sin{(C-B)}}{\sin{2C}}\ge 0$
QUESTION [6 upvotes]: let $0<A,B,C<\dfrac{\pi}{2}$,and $A+B+C=\pi$,prove that
$$\dfrac{\sin{(A-B)}\sin{(A-C)}}{\sin{2A}}+\dfrac{\sin{(B-C)}\sin{(B-A)}}{\sin{2B}}+\dfrac{\sin{(C-A)}\sin{(C-B)}}{\sin{2C}}\ge 0$$
my idea
$$\sin{2B}\sin{2C}\sin{(A-B)}\sin{(A-C)}+\sin{2A}\sin{2C}\sin{(B-C)}\sin{(B-A)}+\sin{2A}\sin{2B}\sin{(C-A)}\sin{(C-B)}\ge 0$$
But following I can't, 
and I find sometime,I have find the Similar quetions:http://www.artofproblemsolving.com/Forum/viewtopic.php?p=128597&sid=55397cc8fe6b896712e5495b124dd247#p128597

REPLY [2 votes]: $$\sin{2B}\sin{2C}\sin{(A-B)}\sin{(A-C)}=\\\sin(2(\pi-A-C))\sin(2(\pi-A-B))\sin{(A-B)}\sin{(A-C)}\\=4[\sin(A+C)\cos(A+C)][\sin(A-C)\cos(A-C)]\sin(A+B)\sin(A-B)\\ =\cos(B)\cos(C)[\sin^{2}(A)-\sin^{2}(B)][\sin^{2}(A)-\sin^{2}(C)]$$
Using the extended Schur's Inequality
 $$\sum \cos(B)\cos(C)[\sin^{2}(A)-\sin^{2}(B)][\sin^{2}(A)-\sin^{2}(C)]\geq0$$