# Thread: proof for an equation

1. ## proof for an equation

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.

2. 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.

3. 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.
$\displaystyle x^2 + y^2 + z^2 = xy + xz + yz$

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

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

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

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

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

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?

4. Originally Posted by Isomorphism
$\displaystyle x^2 + y^2 + z^2 = xy + xz + yz$

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

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

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

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

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

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.

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

6. Originally Posted by Isomorphism
$\displaystyle x^2 + y^2 + z^2 = xy + xz + yz$

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

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

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

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

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

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.