Let $ABC$ be an acute angle triangle.
Show that :
\begin{equation} \sum _{cyc}(\sin2B+\sin2C)^{2} \sin A \leq 12 \sin A \sin B \sin C \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(*)\end{equation}
My try:
I think one can start with applying the addition and subtraction formulas(maybe)! Then he/she has:
\begin{equation}(\sin2B+\sin2C)^2\sin A = 4\sin^2(B+C)\cos^2(B-C)\sin A\end{equation}
Since $A+B+C = 180$ then latest equation equals to:
\begin{equation} = 4\sin^3 A\cos^2(B-C)\end{equation}
But I have no ideas for continue :(
Edited($1/27/2014$) I used labbhattacharjee idea to writing the last equation in new form:
$4\sin^3A\cos(B-C) = 4\sin^2A\sin(B+C)\cos(B-C)$ $$=2\sin^2A(\sin2B + \sin2C)=(1-\cos2A)(\sin2B+\sin2C)$$ $$=(\sin2B+\sin2C)-\sin2B\cos2A-\sin2C\cos2A$$
Then we can write:
$$\sum_{cyc} 4 \sin^3A\cos(B-C)$$
$$=\sum_{cyc}(\sin 2B+\sin 2C) - \sum_{cyc}\sin2B\cos2A - \sum_{cyc}\sin2C\cos2A$$
$$=2\sum_{cyc} \sin2A-\sum_{cyc}\sin2B\cos2A-\sum_{cyc}\sin2A\cos2B$$
$$=2\sum_{cyc} \sin2A - \sum_{cyc}(\sin2B\cos2A+\sin2A\cos2B)$$
$$=2\sum_{cyc}\sin(2A)- \sum_{cyc}\sin(2A+2B)$$
$$=2 \sum_{cyc}\sin2A + \sum_{cyc} \sin2C$$
$$= 3(\sin2A +\sin2B + \sin2C)$$
and by using equality proved in this page we conclude that (look at equation $(*)$):
$$ \sum _{cyc}(\sin2B+\sin2C)^{2} \sin A = 12 \sin A \sin B \sin C \leq 12 \sin A \sin B \sin C $$
and the equality becomes true when $ABC$ is a equilateral.
Is there any other solution?
