<s>[INST] How prove this inequality $x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$ let $x,y,z\ge 0$,and such
$$xy+yz+xz=xyz+2$$
show that
$$x^2+y^2+z^2+xyz(x+y+z-2)\ge 4$$
my try: let $x+y+z=p,xy+yz+xz=q, xyz=r$
then 
$$q=r+2$$
show that
$$p^2-2q+r(p-2)\ge 4$$
then I can't,Thank you
 [/INST] Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, the condition does not depend on $u$ and we need to prove that $f(u)\geq0$,
where $f(u)=9u^2-6v^2+w^3(3u-2)-4$.
But $f$ is an increasing function, which says that it's enough to prove our inequality
for a minimal value of $u$, which happens for equality case of two variables.
Let $y=x$. Then $z=\frac{2-x^2}{x(2-x)}$ and we need to prove that
$$(x-1)^2(x^2-2)^2(2x+1)\geq0,$$
which is obvious.
Done!
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