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 \(\sqrt { x } =\sqrt { y } +\sqrt { z } \).

## Comments

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TopNewestbut 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 方 · 2 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. – Kelvin Hong 方 · 2 months, 1 week ago

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– Steven Jim · 2 months, 1 week 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.Log in to reply

– Kelvin Hong 方 · 2 months, 1 week ago

Yes, so only Case 1 exists.Btw, There isn't any flaw.Log in to reply

– Steven Jim · 2 months, 1 week ago

Thanks!Log in to reply

– Kelvin Hong 方 · 2 months, 1 week ago

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

– Steven Jim · 2 months, 1 week ago

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

– Steven Jim · 2 months, 1 week ago

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

– Kelvin Hong 方 · 2 months, 1 week ago

Okay, I have seen it before, it seems harder... Yeah, I have attempt it .Log in to reply