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I tried to solve this inequality because I find exotic. Actually, I didn't look at the right solution. Because before I look at the right solution, I want to know if my solution is right or not.

Prove that $0 ≤ yz + zx +xy -2xyz≤\frac{7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$

Attempts:

If $x=0$. The left side of the inequality is correct. So, I can accept $x,y,z≠0$.

It is enough to prove $\frac 1x+\frac1y+\frac1z≥2$

We have,

$x+y+z≥3\sqrt[3]{xyz}\Longrightarrow xyz≤\frac1{27}$

$\frac 1x+\frac1y+\frac1z≥\frac3{\sqrt[3]{xyz}}≥9≥2$

The left side proved.

It is obvious at least one of the numbers is less than $\frac 12.$

So, we can choose $y$, such that $y≤\frac 12$.

Therefore, we have

$yz + zx +xy −2xyz ≤ \frac7{27}$

$x(y+z)+yz(1-2x)-\frac7{27}≤0$

$x(1-x)+y(1-x-y)(1-2x)-\frac7{27}≤0$

$x-x^2+y-xy-y^2-2xy+2x^2y+2xy^2-\frac7{27}≤0$

$x^2-x-y+xy+y^2+2xy-2x^2y-2xy^2+\frac{7}{27}≥0$

$x^2(1-2y)+x(3y-2y^2-1)+(y^2-y+\frac7{27})≥0$

$2(\frac12-y)\left(x+\frac{y-1}{2}\right)^2+\frac{1}{108} (3y-1)^2(6y+1)≥0$

Of course, I'm not sure the solution is correct. Can you verify the solution?

Thank you.

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    $\begingroup$ $$(1-x)(1-y)(1-z)=xy+yz+zx-xyz$$ $\endgroup$ – lab bhattacharjee May 30 '19 at 7:45
  • $\begingroup$ Second term of last line is always positive for positive y, but factor 1/2-y in first term can be negative , if y>1/2, what prevents first term from overtaking second terms positivity? $\endgroup$ – mathreadler May 30 '19 at 7:50
  • $\begingroup$ @mathreadler yes this is a critical Step. We can always choose $y$, such that $y≤1/2$ $\endgroup$ – lone student May 30 '19 at 8:16
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    $\begingroup$ @Elvin ah you are correct, I missed this line in the middle where you showed that. Then it should be correct I think. $\endgroup$ – mathreadler May 30 '19 at 8:49
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    $\begingroup$ Looks correct. Simpler approaches exist and may be preferable, though what is simpler is somewhat subjective. I am sure you wouldn’t want to prove the monic cubic whose roots are $x, y, z$ satisfies $p(\frac12) \in [\frac{27}{864}, \frac{251}{864}]$, which is another statement of the same inequality. $\endgroup$ – Macavity May 30 '19 at 12:18
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I checked your solution. Your solution is right.

I like the following way.

The homogenization helps.

By AM-GM $$xy+xz+yz-2xyz=(x+y+z)(xy+xz+yz)-2xyz\geq9xyz-2xyz=7xyz\geq0.$$ Also, $$xy+xz+yz-2xyz\leq\frac{7}{27}$$ it's $$(xy+xz+yz)(x+y+z)-2xyz\leq\frac{7}{27}(x+y+z)^3$$ or $$\sum_{cyc}(7x^3-6x^2y-6x^2z+5xyz)\geq0,$$ which is true by Schur and AM-GM.

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    $\begingroup$ Please, look my Question and tags correctly. $\endgroup$ – lone student May 30 '19 at 7:48
  • $\begingroup$ @Elvin I saw. I did't like it. Sorry. I'll try to find a mistake. $\endgroup$ – Michael Rozenberg May 30 '19 at 7:50
  • $\begingroup$ @Elvin I checked your solution. Your solution is true. $\endgroup$ – Michael Rozenberg May 30 '19 at 7:57
  • $\begingroup$ You could use the Karus-Kuhn-Tucker conditions for optimality to find the maximum of that expression with the restrictions. $\endgroup$ – J. P. C. Jan 13 at 13:46
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Assume $x= \max\{\,x,\,y,\,z\,\}\,\therefore\,3\,x\geqq 1$.

For $z= 1- x- y$, we need to prove

$$x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}\geqq 0$$

We have

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \left \{ 4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2} \right \}+$$

$$+ \frac{1}{3}(\,2\,x- 1\,)(\,3\,y- 1\,)(\,3\,x+ 3\,y- 2\,)$$

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \frac{1}{27}(\,3\,x- 1\,)^{\,2}- \frac{1}{9}(\,2\,x- 1\,)(\,3\,y- 1\,)(\,3\,x+ 3\,y- 2\,)$$

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \frac{1}{108}(\,1- 3\,x\,)^{\,2}(\,6\,x+ 1\,)+ \frac{1}{4}(\,1- 2\,x\,)(\,x+ 2\,y- 1\,)^{\,2}$$

Because

$$4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2}\geqq 0$$

$$\because\,{\rm discriminant}[\,4(\,1- 2\,x\,)y^{\,2}- 4(\,x- 1\,)(\,2\,x- 1\,)y+ \frac{1}{27}(\,5- 9\,x\,)^{\,2},\,y\,]=$$

$$= \frac{32}{27}(\,2\,x- 1\,)(\,3\,x- 1\,)^{\,3}\leqq 0$$

So

$$x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}\geqq 0$$

$\lceil$ ANOTHER!way $\rfloor$

As @Elvin's above, we can choose $1- 2\,x\geqq 0$ (or $\because\,(\,1- 2\,x\,)(\,1- 2\,y)\leqq 0\,\because\,x+ y< 1$).

$$\because\,x^{\,2}+ y^{\,2}+ 3\,xy- 2\,x^{\,2}y- 2\,y^{\,2}x- x- y+ \frac{7}{27}=$$

$$= \frac{1}{108}(\,1- 3\,x\,)^{\,2}(\,6\,x+ 1\,)+ \frac{1}{4}(\,1- 2\,x\,)(\,x+ 2\,y- 1\,)^{\,2}\geqq 0$$

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