# 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!

• Isn't the question obvious? – user35508 Mar 11 '17 at 9:42
• @user35508 Why it's obvious? – Michael Rozenberg Mar 11 '17 at 9:44
• Let me post an answer..Then tell me your views.. – user35508 Mar 11 '17 at 9:44
• Sorry..I missed a point – user35508 Mar 11 '17 at 9:53

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.