<s>[INST] 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
 [/INST] 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.
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