Proving x^2+y^2+z^2 is ≥3 Given x+y+z+xy+yz+zx=6

[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.

 

[Dragonfall]

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

 

[arildno]

Well, let's see:

Rewrite your equation as:
(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?

 

[mrandersdk]

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

 

[arildno]

That was a much better idea!

 

[John Creighto]

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

 

[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.

 

[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.