Not sure if I got this right

I've found this out when I was solving for a problem. Not sure though if I'm correct, so can anyone help me check for any flaws?

"$x, y, z$ are positive real numbers such that $x^2+y^2+z^2=2(xy+yz+zx)$."

Assume that $x\ge y\ge z$

$=>{ x }^{ 2 }+{ (y+z) }^{ 2 }=2(xy+xz)+4yz$

$=>{ (x-y-z) }^{ 2 }=4yz$

Case 1: $x=y+z+2\sqrt { yz }$

$=>\sqrt { x } =\sqrt { y } +\sqrt { z }$

Case 2: $y+z=x+2\sqrt { yz }$

${ (\sqrt { y } -\sqrt { z } ) }^{ 2 }=x$

$=>\sqrt { z } =\sqrt { y } +\sqrt { x }$ or $=>\sqrt { y } =\sqrt { x } +\sqrt { z }$

Of course, $\sqrt { x }\ge \sqrt { y }\ge \sqrt { z }$ and $x, y, z$ are positive real numbers, so it is only possible that $\sqrt { x } =\sqrt { y } +\sqrt { z }$ .

#Algebra

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Comments

but I not sure if Case 2 exixts, because $\sqrt{y}=\sqrt{x}+\sqrt{z}$ ,so $\sqrt{y} > \sqrt{x}$ , $y > x$ which is a contradiction.

Kelvin Hong - 4 years ago

Because $y \geq z$ , so $\sqrt{y} \geq \sqrt{z}$ , so $\sqrt{y}=\sqrt{x}+\sqrt{z}$ is correct.

Kelvin Hong - 4 years ago

Well, that's why I also have to mention that $x,y,z$ are all positive reals. If $x,y,z$ are non-negative reals, then case 2 is also true.

Steven Jim - 4 years ago

Yes, so only Case 1 exists.Btw, There isn't any flaw.

Kelvin Hong - 4 years ago

@Kelvin Hong – Thanks!

Steven Jim - 4 years ago

@Steven Jim – You're welcome. So for the question "Pretty straightfoward", can you post your solution in that page?

Kelvin Hong - 4 years ago

@Kelvin Hong – I will, soon. So you are the only one who attempt the problem, right?

Steven Jim - 4 years ago

@Steven Jim – Okay, I have seen it before, it seems harder... Yeah, I have attempt it .

Kelvin Hong - 4 years ago

@Kelvin Hong – By the way, go ahead and try "Inspired by Khang Nguyen Thanh". I think it is a good one.

Steven Jim - 4 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$