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 $(1)$

We can assume that $x,y,z$ are sides of a triangle because otherwise the LHS is negative.
By Heron's formula we have
$$A^2 = s(s-x)(s-y)(s-z) = \frac1{16}(x+y+z)(x+y-z)(y+z-x)(x+z-y)$$
where $A$ is the area and $s$ is the semiperimeter.
Now recall this inequality:
$$4\sqrt{3}A \le \frac{9xyz}{x+y+z}$$
or $$A^2 \le \frac{27}{16} \frac{(xyz)^2}{(x+y+z)^2}$$
It is listed on wikipedia and for a proof see here.
We get
\begin{align}
(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3 = 16{A^2}(x+y+z)^2 \le {27(xyz)^2}
\end{align}
