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

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In summary, the conversation discusses ways to prove that x^2+y^2+z^2 is greater than or equal to 3 given the equation x+y+z+xy+yz+zx=6. Suggestions include using Lagrange multipliers and Cauchy's inequality.
  • #1
pixel01
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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.
 
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  • #2
What are x, y, z? Real numbers? Positive reals?
 
  • #3
Well, let's see:

Rewrite your equation as:
[tex](x,y,z)\cdot(1+y,1+z,1+x)=6\to||(x,y,z)||||(1+y,1+z,1+x)||\cos\phi=6[/tex]
whereby is implied:
[tex](x^{2}+y^{2}+z^{2})\sqrt{1+\frac{2x+2y+2z+3}{x^{2}+y^{2}+z^{2}}}\geq{6}[/tex]
Maybe this is usable?
 
  • #4
  • #5
That was a much better idea!
 
  • #6
mrandersdk said:
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
 
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  • #7
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
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.
 

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

1. How do you prove that x^2+y^2+z^2 is ≥3?

The proof for this statement involves using algebraic manipulation and the given equation x+y+z+xy+yz+zx=6. By rearranging terms and applying the Cauchy-Schwarz inequality, we can show that x^2+y^2+z^2 is greater than or equal to 3.

2. What is the significance of the inequality x^2+y^2+z^2 ≥3 in this context?

This inequality is significant because it provides a lower bound for the sum of squares of three variables, given a specific condition (x+y+z+xy+yz+zx=6). It also highlights the relationship between the variables x, y, and z in terms of their squares.

3. Can the inequality x^2+y^2+z^2 ≥3 be proven using other methods?

Yes, there are multiple ways to prove this inequality. Some other methods include using the AM-GM inequality, the Power Mean inequality, or geometric proofs.

4. Is there a specific domain or range for the variables x, y, and z in this equation?

The given equation x+y+z+xy+yz+zx=6 does not restrict the domain or range of the variables x, y, and z. However, in order for the inequality x^2+y^2+z^2 ≥3 to hold, the variables must be real numbers.

5. How can this proof be applied in other mathematical contexts?

The concept of proving an inequality using a given equation and algebraic manipulation can be applied in various mathematical contexts, such as optimization problems, number theory, and geometry. This specific proof can also be extended to higher dimensions by using similar techniques.

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