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Homework Help: Another proof: x^2 + xy +y^2 > 0

  1. Sep 17, 2009 #1
    Hello again, I have another proof that I can't figure out how to solve.

    1. The problem statement, all variables and given/known data

    \text{Prove that if }\textit{x }\text{and }\textit{y }\text{are not both 0, then}

    2. Relevant equations


    3. The attempt at a solution

    Not sure if I'm on the right track here...

    \text{Assume (1) is true:}
    x^2+xy+y^2 &> 0\\
    x^2 +2xy + y^2 &> xy\\
    (x+y)^2 &> xy\\

    ...not sure where to go from here.
  2. jcsd
  3. Sep 17, 2009 #2


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    I would say to look for the extrema of f(x,y)=x^2+xy+y^2. Find df/dx and df/dy and set them equal to zero and solve for x and y. Can you classify that as a min or a max or neither?
  4. Sep 17, 2009 #3


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    (x+y)2 = x2 + 2xy + y2 >= 0 You know that already


    x2 + xy + y2 >= -xy

    If x and y are both positive, the result is trivial. If x and y are both negative, the result is also trivial. (in both cases, each term in the summation is positive). When one of x or y is negative, -xy becomes positive. So what can you say?
  5. Sep 17, 2009 #4
    Okay, here's another attempt.

    x^3 - y^3 &= (x-y)(x^2+xy+y^2)\\
    \frac{x^3 - y^3}{x-y} &= x^2+xy+y^2 \tag{1}\\
    x^2 + xy + y^2 &> 0\\
    \frac{x^3 - y^3}{x-y} &> 0 \tag{2}
    \text{(2) is true for }
    \text{, }
    \begin{math}x, y \in \mathbb{R}.\end{math}\\
    \text{For }
    x^2+xy+y^2 &> 0\\
    x^2 + x^2 + x^2 &> 0\\
    3x^2 &> 0\tag{3}
    \text{(3)} is true for all
    \begin{math}x, y \in \mathbb{R}.\end{math}
  6. Sep 17, 2009 #5
    When one of x or y is negative, the LHS will be greater than a positive number and therefore greater than zero?
  7. Sep 17, 2009 #6
    a worse idea for a proof used to live here.
    Last edited: Sep 18, 2009
  8. Sep 17, 2009 #7
    Thank you all!
  9. Sep 18, 2009 #8


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    Well, I think you guys are complicating things a little bit here. :rolleyes:

    We can use the idea of completing squares to solve this problem. It goes like this:

    [tex]x ^ 2 + xy + y ^ 2 = \left[ x ^ 2 + 2 x \left( \frac{1}{2} y \right) + \left( \frac{1}{2}y \right) ^ 2 \right] + \frac{3}{4} y ^ 2[/tex]

    I'm almost spilling out the answer. Since the OP has already had his own solution, this is just another way to tackle the problem.

    So, what left is to determine when the equation holds. :)
  10. Sep 18, 2009 #9
    There is much easier way to do it:



    [tex](x+y/2)^2 + (\frac{y\sqrt{3}}{2})^2 \geq 0 [/tex]
  11. Sep 18, 2009 #10
    could you show your work please?

  12. Sep 18, 2009 #11
    :smile: I've already showed it.

    if a2 [itex]\geq[/itex] 0 and b2 [itex]\geq[/itex] 0 then a2+b2[itex]\geq[/itex] 0
  13. Sep 18, 2009 #12
    never mind, it was just from completing the square from above (I only skimmed through the thread sorry)
  14. Sep 18, 2009 #13

    wow, that's so concise
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