How do I prove this relationship between positive terms of a G.P.? 
$a$, $b$, $c$, and $d$ are positive terms of a G.P. This is the relationship I'm trying to prove:
$$\frac1{ab} + \frac1{cd} > 2 \left(\frac1{bd} + \frac1{ac} - \frac1{ad}\right)$$

This question was listed under the section on Arithmetic, Geometric, and Harmonic Means. So, I tried using those.
$$A = \frac1{ab}$$
$$B = \frac1{cd}$$
$$A.M. = \frac12(A + B) = \frac12\left(\frac1{ab}+\frac1{cd}\right)$$
This gives the correct term on the left side as well as the $2$ on the right side.
But I am completely blank from here. How do I proceed from here to prove this relationship using means?

So $b=ax, c=ax^2$, and $d=ax^3$, and you have to prove: $$\frac1{a^2x} + \frac1{a^2x^5} > 2 (\frac1{a^2x^4} + \frac1{a^2x^2} - \frac1{a^2x^3})$$
or $$1 + \frac1{x^4} > 2 (\frac1{x^3} + \frac1{x} - \frac1{x^2})$$
or $$\boxed{x^4+1>2(x+x^3-x^2)}$$ 
or
$$x^4-2x^3+2x^2-2x+1>0\iff (x^2-x)^2+(x-1)^2>0,$$ which is obviously true.

You can try to prove the boxed inequality also like this, using AM-GM inequality: $$x^4+x^2\geq 2\sqrt{x^4\cdot x^2} = 2x^3$$ and$$x^2+1\geq 2\sqrt{x^2\cdot 1} = 2x$$
so $$x^4+2x^2+x = (x^4+x^2)+(x^2+1)> 2x^3+2x$$ and we are done again.
