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,zx, y, z are positive real numbers such that x2+y2+z2=2(xy+yz+zx)x^2+y^2+z^2=2(xy+yz+zx)."

Assume that xyzx\ge y\ge z

=>x2+(y+z)2=2(xy+xz)+4yz=>{ x }^{ 2 }+{ (y+z) }^{ 2 }=2(xy+xz)+4yz

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

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

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

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

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

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

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

Note by Steven Jim
2 years, 2 months ago

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1 vote

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

Kelvin Hong - 2 years, 2 months ago

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

Kelvin Hong - 2 years, 2 months ago

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

Steven Jim - 2 years, 2 months ago

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

Kelvin Hong - 2 years, 2 months ago

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@Kelvin Hong Thanks!

Steven Jim - 2 years, 2 months ago

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

Kelvin Hong - 2 years, 2 months ago

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

Steven Jim - 2 years, 2 months ago

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

Kelvin Hong - 2 years, 2 months ago

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

Steven Jim - 2 years, 2 months ago

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