Prove $8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$ 
Let $a$, $b$ and $c$ be angles of an acute triangle. Prove that:
  $$8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$$

I tried am-gm, rm-gm rm-am and a couple of other inequalities but I didn't get anywhere.

Let $u$, $v$ and $w$ be lengths-sides of the triangle such that
$\cos{a}=\frac{v^2+w^2-u^2}{2vw}$, $\cos{b}=\frac{u^2+w^2-v^2}{2uw}$, $\cos{c}=\frac{u^2+v^2-w^2}{2uv}$ and
let $u^2+v^2-w^2=r$, $u^2+w^2-v^2=q$, $v^2+w^2-u^2=p$.
Hence, we need to prove that
$$8\prod_{cyc}\left(1-\frac{r^2}{(p+r)(q+r)}\right)\geq\frac{27pqr}{(p+q)(p+r)(q+r)}$$ or
$$8(pq+pr+qr)^3\geq27pqr(p+q)(p+r)(q+r).$$
Let $pq=z$, $pr=y$ and $qr=x$.
Hence, we need to prove that
$$8(x+y+z)^3\geq27(x+y)(x+z)(y+z),$$
which is AM-GM:
$$\left(\frac{\sum\limits_{cyc}(x+y)}{3}\right)^3\geq\prod_{cyc}(x+y).$$
Done!
