Trouble understanding inequality proved using AM-GM inequality I am studying this proof from Secrets in Inequalities Vol 1 using the AM-GM inequality to prove this question from the 1998 IMO Shortlist. However, I'm lost on the very first line of the solution. 

Let $x,y,z$ be positive real numbers such that $xyz =1$. Prove that 
  $$
 \frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+z)(1+x)} + \frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}
$$

Using AM-GM in the following form:
$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}{8} + \frac{1+z}{8} \ge \frac{3x}{4}$. 
Now, this is where I get lost. Where was this expression derived from?
We conclude that
$\sum_{cyc}\frac{x^3}{(1+y)(1+z)} + \frac{1}{4}\sum_{cyc}(1+x) \ge \sum_{cyc}\frac{3x}{4} \rightarrow \sum_{cyc}\frac{x^3}{(1+y)(1+z)} \ge \frac{1}{4}\sum_{cyc}(2x-1) \ge \frac{3}{4}$
The equality holds for $x=y=z=1$

I hope one of the $3$ parts below would help you understand the proof above.
Part 1: the AM-GM ineq is: $\dfrac{x^3}{(1+y)(1+z)}+\dfrac{1+y}{8}+\dfrac{1+z}{8} \geq 3\sqrt[3]{\dfrac{x^3(1+y)(1+z)}{(1+y)(1+z)\cdot 8\cdot 8}} = \dfrac{3x}{4}$.
Part 2: $\sum_{cyclic} \left(\dfrac{1+y}{8}+\dfrac{1+z}{8}\right) = 2\sum_{cyclic} \dfrac{1+x}{8}= \dfrac{1}{4}\sum_{cyclic}(1+x)$
Part 3: $\sum_{cyclic} \dfrac{3x}{4} - \sum_{cyclic} \dfrac{1+x}{4}=\sum_{cyclic} \left(\dfrac{3x}{4}-\dfrac{1+x}{4}\right)=\sum_{cyclic} \dfrac{2x-1}{4}=\dfrac{1}{2}\sum_{cyclic} x - \dfrac{3}{4}\geq \dfrac{1}{2}\cdot 3\sqrt[3]{\prod_{cyclic}x}=\dfrac{3}{2}-\dfrac{3}{4} = \dfrac{3}{4}$
