proof for an equation

**gfreeman75** · April 28th 2008, 03:46 AM

I have this equation here:

x^2 + y^2 + z^2 = xy + xz + yz

I need to somehow prove that the equation is true only when all three variables are equal, i.e., x = y = z.

I would appreciate any advice or suggestions on how to accomplish this.

**mr fantastic** · April 28th 2008, 04:25 AM

Originally Posted by gfreeman75

I have this equation here:

x^2 + y^2 + z^2 = xy + xz + yz

I need to somehow prove that the equation is true only when all three variables are equal, i.e., x = y = z.

I would appreciate any advice or suggestions on how to accomplish this.

Well, I'll probably get jumped on here but here goes .......

The equation can be re-arranged in two ways:

x(x - y) + y(y - z) + z(z - x) = 0 .... (1)

x(x - z) + y(y - x) + z(z - y) = 0 ....(2)

Compare and equate coefficients of x, y and z:

x - y = x - z => y = z.

y - z = y - x => x = z.

Therefore x = y = z is the only solution.

**Isomorphism** · April 28th 2008, 06:32 AM

Originally Posted by gfreeman75

I have this equation here:

x^2 + y^2 + z^2 = xy + xz + yz

I need to somehow prove that the equation is true only when all three variables are equal, i.e., x = y = z.

I would appreciate any advice or suggestions on how to accomplish this.

$x^2 + y^2 + z^2 = xy + xz + yz$

$\Rightarrow 2(x^2 + y^2 + z^2) = 2(xy + xz + yz)$

$\Rightarrow 2(x^2 + y^2 + z^2) - 2(xy + xz + yz) = 0$

$\Rightarrow (x^2 + y^2 - 2xy)+(x^2 + z^2 - 2xz) + (z^2 + y^2 - 2zy) = 0$

$\Rightarrow (x-y)^2+(x-z)^2 + (z-y)^2 = 0$

$\text{This forces }x=y=z.$

Originally Posted by Mr.F

Well, I'll probably get jumped on here but here goes .......

Mr F, can you please explain the validity of your method?

**Jameson** · April 28th 2008, 06:36 AM

Originally Posted by Isomorphism

$x^2 + y^2 + z^2 = xy + xz + yz$

$\Rightarrow 2(x^2 + y^2 + z^2) = 2(xy + xz + yz)$

$\Rightarrow 2(x^2 + y^2 + z^2) - 2(xy + xz + yz) = 0$

$\Rightarrow (x^2 + y^2 - 2xy)+(x^2 + z^2 - 2xz) + (z^2 + y^2 - 2zy) = 0$

$\Rightarrow (x-y)^2+(x-z)^2 + (z-y)^2 = 0$

$\text{This forces }x=y=z.$

Mr F, can you please explain the validity of your method?

Very succinct. It took me at least twice as many lines when I tried working it out.

**gfreeman75** · April 28th 2008, 09:38 AM

That's a pretty elegant solution to my problem, Isomorphism. Thank you.

**mr fantastic** · April 28th 2008, 01:08 PM

Originally Posted by Isomorphism

$x^2 + y^2 + z^2 = xy + xz + yz$

$\Rightarrow 2(x^2 + y^2 + z^2) = 2(xy + xz + yz)$

$\Rightarrow 2(x^2 + y^2 + z^2) - 2(xy + xz + yz) = 0$

$\Rightarrow (x^2 + y^2 - 2xy)+(x^2 + z^2 - 2xz) + (z^2 + y^2 - 2zy) = 0$

$\Rightarrow (x-y)^2+(x-z)^2 + (z-y)^2 = 0$

$\text{This forces }x=y=z.$

Mr F, can you please explain the validity of your method?

Well, I guess I'm saying that if you've got

ax + by + cz = 0 .... (1)

cx + ay + bz = 0 ....(2)

then a = b = c if x, y and z are not equal to zero.

A bit lame, I guess.

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