<s>[INST] In the provided question of AM-GM inequality in the description, I'm not able to figure out which manipulation was used here, in the solution? 
Let $x, y, z \in \Bbb R_{> 0}$ such that $xyz=1$. Prove that $$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}$$


Now, in the first step they suggested that we can use AM-GM in the following form
$$\frac{x^3}{(1+y)(1+z)} + \frac{(1+y)}{8} + \frac{(1+z)}{8} \ge \frac{3x}{4}$$
I am wondering what manipulation they did in the first step to write the following expression as stated above? A little help is appreciated.
 [/INST] The expression wasn't rewritten, those other two terms were more or less poofed into existence. Implicitly we're saying
$$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)}$$ $$ = \frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} + \frac{(1+y)}{8} + \frac{(1+z)}{8} - \frac{(1+y)}{8} - \frac{(1+z)}{8}$$
Why does that help? Because cyclicly adding and subtracting those two terms and then applying AM-GM on all three resulting expressions gives
$$\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+x)(1+z)} + \frac{z^3}{(1+x)(1+y)} \geq \frac{3(x+y+z)}{4} - \frac{(3+x+y+z)}{4}$$
$$ = \frac{x+y+z}{2}-\frac{3}{4}$$
Can you finish the last step?
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