# 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 $\sqrt { x } =\sqrt { y } +\sqrt { z }$.

Note by Steven Jim
3 years, 10 months ago

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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.

- 3 years, 10 months ago

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

- 3 years, 10 months 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.

- 3 years, 10 months ago

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

- 3 years, 10 months ago

Thanks!

- 3 years, 10 months ago

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

- 3 years, 10 months ago

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

- 3 years, 10 months ago

Okay, I have seen it before, it seems harder... Yeah, I have attempt it .

- 3 years, 10 months ago

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

- 3 years, 10 months ago