Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality it's
which is a linear inequality of $w^3$,
which says that it's enough to prove our inequality for an extremal value of $w^3$.
Now,we see that $x$, $y$ and $z$ are non-negative roots of the following equation.
which says that the graph of $f(X)=X^3-3uX^2+3v^2X$ and the line $Y=w^3$
have three common points: $(x,f(x))$, $(y,f(y))$ and $(z,f(z))$.
Now, let $u$ and $v^2$ be constants and $w^3$ changes.
We see that $w^3$ gets a maximal value, when the line $Y=w^3$ will touch to the graph of $f$,
which happens for equality case of two variables.
Also, we see that $w^3$ gets a minimal value, when the line $Y=w^3$ will touch to the graph of $f$,
which happens for equality case of two variables, or when $w^3=0$.
Thus, it's enough to prove our inequality in the following cases.
Hence, we need to prove here that
which is C-S and AM-GM:
We need to prove
which is AM-GM and C-S: