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# 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
7 months, 1 week ago

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## Comments

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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.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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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.

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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Thanks!

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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

- 7 months, 1 week ago

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