# A small problem

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.

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

arildno
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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?

arildno
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That was a much better idea!!

maybe use lagrange multiplier to find the extremums of x^2+y^2+z^2 under the constraint x+y+z+xy+yz+zx=6, then verify that the minimum gives the value 3.

see:

http://en.wikipedia.org/wiki/Lagrange_multipliers

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

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

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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.

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.