# An Inequality with Sines

### Solution

We'll use several known facts:

$\displaystyle \sin^2x=\frac{1-\cos 2x}{2},\\ \displaystyle 1+\sum_{cycl}\cos 4A=4\cos 2A\cos 2B\cos 2C,\\ \displaystyle -1-\sum_{cycl}\cos 2A =4\cos A\cos B\cos C,\\ \displaystyle \cos A\cos B\cos C\le\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}.$

We first observe that the required inequality is equivalent to

$-\cos 2A\cos 2B\cos 2C\le\cos A\cos B\cos C,$

which, in turn, is equivalent to

$\displaystyle\cos (\pi-2A)\cos(\pi-2B)\cos(\pi-2C)\le\sin\frac{\pi-2A}{2}+\sin\frac{\pi-2B}{2}+\sin\frac{\pi-2C}{2},$

which is true, following $\displaystyle \cos A\cos B\cos C\le\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}.$

### Acknowledgment

I am grateful to Leo Giugiuc for communicating to me Nguyen Viet Hung's elegant problem, along with a solution of his.