Prove: $\log_2(x)+\log_x(y)+\log_y(8)\geq \sqrt[3]{81}$ 
Prove that for every $x$,$y$ greater than $1$:
  $$\log_2(x)+\log_x(y)+\log_y(8)\geq \sqrt[3]{81}$$

What I've tried has got me to:
$$\frac{\log_y(x)}{\log_y(2)}+\log_x(y)+3\log_y(2)\geq \sqrt[3]{81}$$
I didn't really get far.. I can't see where I can go from here, especially not what to do with $ \sqrt[3]{81}$.
This is taken out of the maths entry tests for TAU, so this shouldn't be too hard.

Two things:
$\log_a b = \frac 1{\log_b a}$ and $\frac {\log_b c}{\log_b a} = \log_a c$ so $\log_a b\log_b c = \frac {\log_b c}{\log_b a} = \log_a c$.
And AM-GM says $\frac {a  + b+ c}3 \ge \sqrt[3]{abc}$.
So.....
$\frac {\log_2 x + \log_x y + \log_y 8}3 \ge \sqrt[3]{\log_2 x \log_x y \log_y 8} = \sqrt[3]{\log_2 8}$
