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

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

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

arildno
Homework Helper
Gold Member
Dearly Missed
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.

Mark44
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.

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

--Elucidus

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.

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.

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.