Trouble understanding inequality proved using AM-GM inequality

Question

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 $x=y=z=1$

Answer 1

Let me try to cover the motivation for writing something like $$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}8 + \frac{1+z}8$$ for doing AM-GM, seemingly out of the blue.

We obviously start with the first term. Now clearly, to get rid of the denominator, it would be great to consider the terms $(1+y)$ and $(1+z)$. As there is $x^3$ in the numerator, we can use these as separate terms to do a three term AM-GM, to get a simple $x$ term after the AM-GM. An equally good alternative would be to use the single term $(1+y)(1+z)$ and do a two term AM-GM. So far so good.

Now note that AM-GM will have equality iff all terms are equal. However we know equality is when $x=y=z=1$. We have to maintain this equality possibility in each step taken in the proof, otherwise it cannot hold finally. This leads immediately to having the coefficient $\frac18$ attached to the second and third terms, so that the value of all three terms are the same, viz. $\frac14$ when $x=y=z=1$. Now you have the right set of terms to do the AM-GM!

Answer 2

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 $$a=\frac{x^3}{(1+y)(1+z)}\ ,\quad b=\frac{1+y}8\ ,\quad c=\frac{1+z}8\ .$$

Answer 3

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: $\sum_{cyclic} \dfrac{3x}{4} - \sum_{cyclic} \dfrac{1+x}{4}=\sum_{cyclic} \left(\dfrac{3x}{4}-\dfrac{1+x}{4}\right)=\sum_{cyclic} \dfrac{2x-1}{4}=\dfrac{1}{2}\sum_{cyclic} x - \dfrac{3}{4}\geq \dfrac{1}{2}\cdot 3\sqrt[3]{\prod_{cyclic}x}=\dfrac{3}{2}-\dfrac{3}{4} = \dfrac{3}{4}$