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.
 A: 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!
A: Here's a proof that accounts for all acute-angled triangles whose angles lie in $ \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$.
Note that $f(x) = \ln \sin(x) \tan(x)$ is convex over $ \left( \tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$ since $f''(x) = \dfrac{\tan^{2}(x) -2}{\sin^{2}(x)} > 0 \quad \forall x\in \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$
Hence we can apply Jensen's inequality to $f(x)$ and $a, b,c \in \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$ to get:
$f\left(\dfrac{a+b+c}{3}\right)\leq \dfrac{f(a) + f(b)+ f(c)}{3}$
$\Rightarrow \ln \dfrac{3}{2} \leq \dfrac{\ln \dfrac{\sin^{2}(a)\,\sin^{2}(b)\,\sin^{2}(c)}{\cos(a)\,\cos(b)\,\cos(c)}}{3}$
$\Rightarrow \dfrac{27}{8} \leq \dfrac{\sin^{2}(a)\,\sin^{2}(b)\,\sin^{2}(c)}{\cos(a)\,\cos(b)\,\cos(c)}$, as required.
Can someone figure out an extension of this argument that accounts for those acute-angled triangles with one angle in $ \left(0, \tan^{-1}{\sqrt{2}}\right)$?
