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 $(1)$
 A: 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}
A: 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:
$$4x^2y^2z^2\geq(x+y-z)(x+z-y)(y+z-x)(x^3+y^3+z^3+xyz).$$
A: $$\color{brown}{\textbf{Final edition (13.08.18)}}$$
$\color{green}{\textbf{Task transformations.}}$
The problem is homogeneous with respect to unknowns $x,y,z.$
Let WLOG
$$x+y+z=6,\tag1$$
Then the equivalent inequality is
$$x^2y^2(6-x-y)^2\ge64(3-x)(3-y)(x+y-3),\tag2$$
where
$$(x,y,6-x-y)\in(0,6),$$
or 
$$((3-y)+(x+y-3),(3-x)+(x+y-3),(3-x)+(3-y))\in(0,6).\tag3$$
Inequality $(2)$ is satisfied if one or three of the factors $(3-x),(3-x),(x+y-3)$ are non-positive.
On the other hand, conditions $(3)$ is not satisfied if two of this tactors are non-positive. 
Thus, it remains to prove the inequality $(2)$ in the case
$$(x,y,x+y-3)\in(0,3)^3.\tag4$$
Denote
$$t= xy,\tag5$$
then
$$(3-x)(3-y)= xy - 3(x+y) +9 = t-3(6-z)+9 = t+3z-9,$$
$$t=xy\le\left(\frac{x+y}2\right)^2=\frac{(6-z)^2}4,$$
and the stronger problem is to prove the inequality
$$z^2t^2\ge 64(t+3z-9)(3-z),$$
or
$$f(z,t)=z^2t^2+64(z-3)t + 192(z-3)^2\ge0,\tag6$$
under the conditions
$$z\in(0,3),\quad t\in\left(0,\frac{(6-z)^2}4\right].\tag7$$
$\color{green}{\textbf{Proof.}}$
The least value of $f(z,p)$ can be achieved only in the stationary points or in the bounds.
The stationary points of $f(z,p)$ can be found from the system $f'z=f'_t = 0,$ or
\begin{cases}
zt^2+32t + 192(z-3) = 0\\
z^2t+32(z-3) = 0.
\end{cases}
Taking in account $(7),$ one can obtain
$$\begin{cases}
(zt+32-6z^2)t = 0\\
z^2t+32(z-3) = 0,
\end{cases}\rightarrow
\begin{cases}
zt=6z^2-32\\
z(6z^2-32)+32(z-3) = 0,
\end{cases}\rightarrow
\begin{cases}
z_s=2\sqrt[3]2\\
t_s=12\sqrt[3]2-8\sqrt[3]4
\end{cases}\\[8pt]
f(z_s,t_s) = 64(11-54\sqrt[3]2+36\sqrt[3]4)\approx7.085>0.
$$
Let us check the bounds.
Case $\mathbf{z\to 0,\quad t\in(0,9).}$
$$f(0, t) = 192(9 - t)>0.$$
Case $\mathbf{z\to 3, quad t > 0.}$
$$f(3, t) = 3t^2 >0.$$
Case $\mathbf{t=\frac{(6-z)^2}4,\quad z\in(0,3).}$
$$f\left(z, \frac{(6-z)^2}4\right) = \frac1{16}z^2(6-z)^4+16(z-3)(6-z)^2 + 192(z-3)^2 = \frac1{16}z^2(z-2)^2(z^2-20z+132)\ge0$$
(see also Wolfram Alpha).
$\textbf{Proved.}$
