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

[suhasm]

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

I tried a lot, but can't get any answer. Can someone please help me out?

 

[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

 

[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.

 

[Mark44]

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.

 

[Elucidus]

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

--Elucidus

 

[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.

 

[geelpheels]

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