As $\sin A\sin B=\cos C+\cos A\cos B,\,$ we need to prove that

$\displaystyle 3\sum_{cycl}\cos A\ge 2\sum_{cycl}\cos A+2\sum_{cycl}\cos A\cos B,$

which is the same as

$\displaystyle \sum_{cycl}\cos A\ge 2\sum_{cycl}\cos A\cos B.$

But

$\displaystyle 1\lt \sum_{cycl}\cos A\le \frac{3}{2}$

so that

$\displaystyle \sum_{cycl}\cos A\ge \frac{2}{3}\left(\sum_{cycl}\cos A\right)^2.$

Suffice it to prove that

$\displaystyle \frac{2}{3}\left(\sum_{cycl}\cos A\right)^2\ge 2\sum_{cycl}\cos A\cos B.$

But this is equivalent, due to, say, the Rearrangement inequality, to the obvious

$\displaystyle \left(\sum_{cycl}\cos A\right)^2\ge 3\sum_{cycl}\cos A\cos B.$

The problem, with a solution, was kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc, who shares credits Adil Abdulayev.