For $2x+2y-3z\geq0$ and similar prove that $\sum\limits_{cyc}(7z-3x-3y)(x-y)^2\geq0$ 
Let $x$, $y$ and $z$ be non-negative numbers such that 
  $2x+2y-3z\geq0$, $2x+2z-3y\geq0$ and $2y+2z-3x\geq0$. Prove that:
  $$(7z-3x-3y)(x-y)^2+(7y-3x-3z)(x-z)^2+(7x-3y-3z)(y-z)^2\geq0$$

I have a proof for the following weaker inequality. 
Let $x$, $y$ and $z$ be non-negative numbers such that
$2x+2y-3z\geq0$, $2x+2z-3y\geq0$ and $2y+2z-3x\geq0$. Prove that:
$$(11z-4x-4y)(x-y)^2+(11y-4x-4z)(x-z)^2+(11x-4y-4z)(y-z)^2\geq0.$$
For the proof we can use the following lemma.
Let $a$, $b$, $c$, $x$, $y$ and $z$ be real numbers such that
$x+y+z\geq0$ and $xy+xz+yz\geq0$. Prove that:
$$(a-b)^2z+(a-c)^2y+(b-c)^2x\geq0.$$
Proof.
Since $x+y+z\geq0$, we see that $x+y$ or $x+z$ or $y+z$ is non-negative because if
$x+y<0$, $x+z<0$ and $y+z<0$ so $x+y+z<0$, which is contradiction.
Let $x+y\geq0$.
If $x+y=0$ so $xy+xz+yz=-x^2\geq0$, which gives $x=y=0$ and since $x+y+z\geq0$, 
we obtain $z\geq0$, which gives $(a-b)^2z+(a-c)^2y+(b-c)^2x\geq0.$
Id est, we can assume $x+y\geq0$.
Now, $$(a-b)^2z+(a-c)^2y+(b-c)^2x=(a-b)^2z+(a-b+b-c)^2y+(b-c)^2x=$$
$$=(x+y)(b-c)^2+2(a-b)(b-c)y+(y+z)(a-b)^2$$
and since $x+y>0$, it's enough to prove that
$$y^2-(x+y)(y+z)\leq0,$$
which is
$xy+xz+yz\geq0,$
which ends a proof of the lemma.
Now we can prove a weaker problem.
From the condition we have:
$$\sum_{cyc}(2x+2y-3z)(2x+2z-3y)=\sum_{cyc}(9xy-8x^2)\geq0$$
and we see that $\sum\limits_{cyc}(11z-3x-3y)=5(x+y+z)\geq0$ and
$\sum\limits_{cyc}(11z-3x-3y)(11y-3x-3z)=9\sum\limits_{cyc}(9xy-8x^2)\geq0$
and by the lemma we are done!
This way does not help for the starting inequality.
Any hint please. Thank you!
 A: Proof: 
Because both the conditions and the conclusion are homogeneous, we may assume that $x+y+z=1$.
Geometrically,
 $$\Delta=\{(x,y,z)\in \mathbb{R}^3 |\ 2x+2y−3z≥0,  2x+2z−3y≥0, \\
2y+2z−3x≥0, x+y+z=1\}$$
 is a triangle . By solveing three linear equations 
$$2x+2y−3z=0,\  2x+2z−3y=0,\ x+y+z=1;$$
$$2x+2z−3y=0,\ 2y+2z−3x=0,\ x+y+z=1;$$
$$2x+2y−3z=0,\  2y+2z−3x=0,\ x+y+z=1;$$
we get three vertices of the triangle $\Delta$. They are $(\frac{1}{5}, \frac{2}{5}, \frac{2}{5})$, $(\frac{2}{5}, \frac{1}{5}, \frac{2}{5})$ and $(\frac{2}{5}, \frac{2}{5}, \frac{1}{5})$.
Then $\Delta=u(\frac{1}{5}, \frac{2}{5}, \frac{2}{5})+v(\frac{2}{5}, \frac{1}{5}, \frac{2}{5})+w(\frac{2}{5}, \frac{2}{5}, \frac{1}{5}),\ u+v+w=1, u\geq 0,v\geq 0,w\geq 0$.
Let $x=\frac{1}{5}u+\frac{2}{5}v+\frac{2}{5}w,\ y=\frac{2}{5}u+\frac{1}{5}v+\frac{2}{5}w,
\ z=\frac{2}{5}u+\frac{2}{5}v+\frac{1}{5}w$. We may write the inequality 
$$(7z−3x−3y)(x−y)^2 +(7y−3x−3z)(x−z)^2 +(7x−3y−3z)(y−z)^2\geq 0$$
as 
$$\dfrac{1}{25}[(u+v-w)(u-v)^2+(v+w-u)(w-v)^2+(u+w-v)(u-w)^2]\geq 0.$$
Obvioursly, it is schur's inequality. Equality holds for $(x,y,z)=(1,1,1)$, and for
$(x,y,z)=(4,3,3)$ or any cyclic permutation.
A: We can use the substitutions $2x+2y - 3z = u, \ 2y+2z - 3x = v$ and $2z+2x - 3y = w$ for $u, v, w \ge 0$.
By solving the system of equations above, we get
$x = \frac{2u+v+2w}{5}, \ y = \frac{2u+2v+w}{5}$ and $z = \frac{u+2v+2w}{5}$.
The rest is the same as @yao4015's solution.
