prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$ Let $x,y,z>0$,show that
$$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$
I have prove this inequality
$$(x+y-z)(y+z-x)(x+y-z)\le xyz$$
because it is three schur inequality
$$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
how to solve this inequiality $xyz ⩾ (x+y-z)(y+z-x)(z+x-y)$
But I can't prove this inequality to prove $(1)$

It's enough to prove our inequality for $\prod\limits_{cyc}(x+y-z)\geq0.$
Now, if $x+y-z\leq0$ and $x+z-y\leq0$ then $x+y-z+x+z-y\leq0$ or $2x\leq0$,
which is a contradiction.
Thus, we can assume that $x+y-z>0$, $x+z-y>0$ and $y+z-x>0.$
Now, let $x+y-z=c$, $x+z-y=b$ and $y+z-x=a$.
Thus, we need to prove prove that
$$27(a+b)^2(a+c)^2(b+c)^2\geq64abc(a+b+c)^3$$ and since
$$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's just $$\sum_{cyc}c(a-b)^2\geq0,$$ it's enough to prove that
$$3(ab+ac+bc)(a+b)(a+c)(b+c)\geq8abc(a+b+c)^2$$ or
$$\sum_{cyc}(3a^3b^2+3a^3c^2-2a^3bc-4a^2b^2c)\geq0,$$
which is true by Muirhead.
A bit of easier way it's to use $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ for the all term $\prod\limits_{cyc}(a+b)^2.$
If so, it's enough to prove that $$(ab+ac+bc)^2\geq3abc(a+b+c),$$ which is obvious. 
The following much more stronger inequality is also true.
Let $x$, $y$ and $z$ be real numbers. Prove that:
$$4x^2y^2z^2\geq(x+y-z)(x+z-y)(y+z-x)(x^3+y^3+z^3+xyz).$$
