<s>[INST] Show that $x^2 + xy^2+ xyz^2\geq 4xyz-4 $ for positive real $x,y,z$ Let $x$,$y$ and $z$ are three positive real numbers.Show that $x^2 + xy^2+ xyz^2\geq 4xyz-4 $.
I have tried to attack the problem by order relationship $x \geq y \geq z$ and then converting them into single variable , AM-GM on $x^2 + xy^2+ xyz^2$  and $x^2 + xy^2+ xyz^2 + 4 $ but nothing is taking me anywhere near to the solution. Please help.
 [/INST] HINT
Note that 
$$x^2+xy^2+xyz^2+4=\frac{x^2}{4}+\frac{x^2}{4}+\frac{x^2}{4}+\frac{x^2}{4}+\frac{xy^2}{8}+\dots+\frac{xy^2}{8}+\frac{xyz^2}{16}+\dots+\frac{xyz^2}{16}+1+1+1+1$$
You can apply AM-GM.
Side Note
Note your attempt to prove the inequality by assuming $x \ge y \ge z$ is wrong as $x,y,z$ are not interchangeable. 
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