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 $x=y=z=1$

The AM-GM inequality for three terms can be written
$$a+b+c\ge3\sqrt[3]{abc}\ .$$
Just substitute your three terms on the LHS into this.  That is, take
$$a=\frac{x^3}{(1+y)(1+z)}\ ,\quad b=\frac{1+y}8\ ,\quad c=\frac{1+z}8\ .$$
