Mathematicals inequalities For $$x,y,z>0 $$
Prove that
$$(2xyz)^2 \ge (x^3+y^3+z^3+xyz)(x+y-z)(y+z-x)(z+x-y)$$
I have tried a famous inequalities: $$(x+y-z)(y+z-x)(z+x-y) \le xyz$$
So the problem is:
$$3xyz \ge x^3+y^3+z^3$$
But in fact, this isn’t true
Help me plz
 A: If $\prod\limits_{cyc}(x+y-z)<0$ so the inequality is obvious.
Thus, it's enough to assume that $\prod\limits_{cyc}(x+y-z)\geq0.$
Now, since $x+y-z<0$ and $x+z-y<0$ gives $x<0$, which is a contradiction, we can assume that
$x+y-z=c\geq0,$ $x+z-y=b\geq0$ and $y+z-x=a\geq0$ and we need to prove that
$$(a+b)^2(a+c)^2(b+c)^2\geq4abc(abc+\sum_{cyc}(a^3+2a^2b+2a^2c))$$ or
$$\sum_{sym}(a^4b^2-a^4bc+a^3b^3-2a^3bc+a^2b^2c^2)\geq0$$ or
$$\prod_{cyc}(a-b)^2+4\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0,$$ which is true by Schur.
A: Suppose $z=\min\{x,y,z\}.$ We can write this inequality as
$$(x^3+y^3+z^3+xyz)\left [xyz-\prod (x+y-z)  \right ] \geqslant xyz(x^3+y^3+z^3-3xyz),$$
But
$$xyz-\prod (x+y-z) = (x+y-z)(x-y)^2+z(x-z)(y-z),$$
$$x^3+y^3+z^3-3xyz = (x+y+z)[(x-y)^2+(x-z)(y-z)].$$
Thefore the inequality equivalent to
$$M(x-y)^2+N(x-z)(y-z) \geqslant 0,$$
for
$$\begin{aligned}M &= (x^3+y^3+z^3+xyz)(x+y-z)-xyz(x+y+z) \\& \geqslant (x^3+y^3+z^3+xyz)x-xyz(x+y+z) \\& = x[x^3+(y+z)(y-z)^2] \geqslant 0.\end{aligned}$$
$$N = x^3+y^3+z^3-xy(x+y) = (x+y)(x-y)^2+z^3 \geqslant 0.$$
The proof is completed.
