# A small problem

1. Feb 17, 2008

### pixel01

I do not know if this is suitable here:

Given that : x+y+z+xy+yz+zx=6,
Prove that x^2+y^2+z^2 >=3

Any hints will be appreciated. Thanks.

2. Feb 17, 2008

### Dragonfall

What are x, y, z? Real numbers? Positive reals?

3. Feb 17, 2008

### arildno

Well, let's see:

$$(x,y,z)\cdot(1+y,1+z,1+x)=6\to||(x,y,z)||||(1+y,1+z,1+x)||\cos\phi=6$$
whereby is implied:
$$(x^{2}+y^{2}+z^{2})\sqrt{1+\frac{2x+2y+2z+3}{x^{2}+y^{2}+z^{2}}}\geq{6}$$
Maybe this is usable?

4. Feb 17, 2008

### mrandersdk

5. Feb 17, 2008

### arildno

That was a much better idea!!

6. Feb 17, 2008

### John Creighto

I'll have to read up on that method. ANother alternative:

http://www.imf.au.dk/da/matematiklaererdag/2005/filer/niels.pdf [Broken]

Last edited by a moderator: May 3, 2017
7. Feb 17, 2008

### mrandersdk

that is a funny coincidens John Creighto i'm studying physics and mathematics at the university of Aarhus denmark, where your note is from.Haven't stydied the note though.

8. Feb 18, 2008

### pixel01

Thanks for all of your contributions.
This one can be solved using Cauchy though.
We have: x^2 + 1 >= 2x; y^2+1>=2y and z^2+1>=2z.
Then it's easy for the rest.