Non-Negative Reals and Inequality: A Proof of xy+yz+zx-xyz≤2

[anemone]

Let $x,\,y,\,z$ be non-negative reals such that $x^2+y^2+z^2+xyz=4$.

Prove that $0\le xy+yz+zx-xyz\le 2$.

 

[Kyle1]

Spoiler

By Cauchy inequality, we have $4=x^2+y^2+z^2+xyz \ge 4\sqrt{x^3 y^3 x^3}$
so $xyz \le 1$ (1)
Then, $xy+yz+zx-xyz \le x^2 +y^2 +z^2 -1$ (since (1) and $x^2+y^2+z^2 \ge xy+yz+zy$) (I think you already know how to prove it)
Thus, $xy+yz+zx-xyz \le 4-xyz-1 \le 4-1-1 =2$
So the right part of the given inequality is proved. But I still have no idea how to prove the other part...:( Can anyone continue ?

 

[anemone]

By Cauchy inequality, we have $4=x^2+y^2+z^2+xyz \ge 4\sqrt{x^3 y^3 x^3}$
so $xyz \le 1$ (1)
Then, $xy+yz+zx-xyz \le x^2 +y^2 +z^2 -1$ (since (1) and $x^2+y^2+z^2 \ge xy+yz+zy$) (I think you already know how to prove it)
Thus, $xy+yz+zx-xyz \le 4-xyz-1 \le 4-1-1 =2$
So the right part of the given inequality is proved. But I still have no idea how to prove the other part...:( Can anyone continue ?

Hi Kyle,

Thank you for participating and partially proving correct of the given inequality. Also, I believe you meant by using the AM-GM inequality to imply that

$\dfrac{x^2+y^2+z^2+xyz}{4} \ge \sqrt[4]{x^3 y^3 x^3}$ and since we're told that $x^2+y^2+z^2+xyz=4$ thus we have

$1 \ge \sqrt[4]{x^3 y^3 x^3}\rightarrow\,\,\therefore xyz \le 1$, am I correct?

I encourage you to think harder on how to prove for the left side inequality and I will be waiting for your answer.

 

[Kyle1]

Oh! Thanks for showing me my mistake. It was AM-GM, not Cauchy inequality haha ;)
The left part of the solution is complex :(

 

[Fallen Angel]

Hi,

A way for the other inequality.

Spoiler

From $xyz\leq 1$ you can assume $x\leq 1$
factor $xy+yz+xz-xyz=y(x+z-xz)+xz$ and now is trivial.

 

[Kyle1]

Hi,

A way for the other inequality.

Spoiler

From $xyz\leq 1$ you can assume $x\leq 1$
factor $xy+yz+xz-xyz=y(x+z-xz)+xz$ and now is trivial.

I don't think it is clear.Can you prove it carefully? If you say so, when xy+yz+zx-xyz is equal to 0? I mean x=? y=? z=?

 

[Fallen Angel]

Hi Kyle,

Spoiler

Without loss of generality you can assume $x\leq z \leq y$ (It's a strange order, but just chosen for fit my post above)
From equation $xyz\leq 1$ you know that at least one of them is less or equal than one, so $x\leq 1$
Now you got $y(x+z-xz)+xz$, and $x+z\geq z \geq xz$ because the statement says that $x,y,z\geq 0$ so both terms are non-negative and so is the sum.
The equality will hold if and only if $x=z=0$

 

[anemone]

Hi Kyle and Fallen Angel,

Sorry for the late reply.:( I will show the other method to prove this inequality, and I am sure both of you and the readers will find it to be another piece of good solution to "read", just like how smart the solution you and Fallen Angel have shown us.

Solution of other:

Spoiler

For the LHS:

Note that we cannot have $x,\,y,\,z>1$. Suppose $z\le 1$. Then we have

$xy+yz+zx-xyz\ge xy+yz+zx-xy=z(x+y)\ge 0$

For the RHS:

By AM-GM inequality, we have $4=x^2+y^2+z^2+xyz\ge 4(xyz)^{\dfrac{3}{4}}\rightarrow\,\,\therefore xyz\le 1$.

By the pigeon hole principle, among three non-negative reals, either two exceed 1 or two are at most 1. Hence we assume

$(x-1)(y-1)\ge 0$, which gives $xy+1\ge x+y$

Thus we have

$xyz+z\ge xz+yz$

$z\ge xz+yz-xyz$

Now, we have

$xy+yz+zx-xyz\ge xy+z$

Either we are done or $xy+z>2$.

But in the latter case, $4=(x^2+y^2)+z(z+2xy)>2xy+2z=2(xy+z)>4$, a contradiction and we're done.