<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] 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.
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