# If x2+y2+z2=xy+yz+zx prove x=y=z

1. Sep 17, 2009

### suhasm

if x2+y2+z2=xy+yz+zx prove that x=y=z.

2. Sep 17, 2009

### Elucidus

In general this is false, unless you know that all three variables are real-valued.

Try:

$$x^2 + (y + z)x + (y^2 + z^2 - yz) = 0$$ and use the quadratic formula and see where that leads.

--Elucidus

3. Sep 17, 2009

### arildno

Rewrite this, for example as:
$$\frac{1}{2}x^{2}+\frac{1}{2}x^{2}+\frac{1}{2}y^{2}+\frac{1}{2}y^{2}+\frac{1}{2}z^{2}+\frac{1}{2}z^{2}-xy-xz-zy=0$$
and see if you can make some clever manipulations.

4. Sep 17, 2009

### Staff: Mentor

And by "clever manipulations," Arildno means group the x2, xy, and y2 terms together, and group the x2, xz, and z2 terms together, and group the y2, yz, and z2 terms together. It's possible that some factorization can occur.

5. Sep 17, 2009

### Elucidus

You will also need $A^2 \geq 0$ for all real A, and equality to 0 occurs only when A = 0.

--Elucidus

6. Sep 17, 2009

### willem2

It is often much easier to prove that something is 0 when squares are involved so I substituted

y = x + a, and z = x+b.

and ended up with

a^2 + b^2 = ab, where you have to prove that a = b = 0.

7. Feb 22, 2011

### geelpheels

A much easier way,

from
x2+y2+z2=xy+yz+zx
you can get
2x2+2y2+2z2=2xy+2yz+2zx
which equals
2x2+2y2+2z2-2xy-2yz-2zx=0
,the equivalant is
(x-y)2+(y-z)2+(z-x)2=0.