1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Show that this inequality is true for all x, y ε R

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    This is part of a question on absolute convergence on series. The following equation is given as a hint. It says that before answering the question on series I should prove that |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R


    2. Relevant equations



    3. The attempt at a solution
    I know that this is the same as |x||y| <= 1/2(|x||x| + |y||y|). That is about all I am able to do to manipulate this equation. I tried solving for x or y but found it is not possible to separate them out. So I can't see how to prove this other than -
    letting x = 0, then showing the equation is true for a few values of y
    letting x = 1, then showing the equation is true for a few values of y
    etc...

    And then saying the equation is true 'by induction'. Surely there is a better, less awkward, way to show that |xy| <= 1/2(|x|^2 + |y|^2) for any x,y ε R
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 2, 2011 #2

    PeterO

    User Avatar
    Homework Helper

    COllect all the terms on one side, and that side will be able to be expressed as a perfect square. I would always multiply each side by 2 to get rid of the nuisance fraction first.
     
  4. Oct 2, 2011 #3
    Ok, if I collect the terms on one side I get
    |x|^2 -2|xy| + |y|^2 >= 0

    which is
    (|x| - |y|)^2 >= 0

    So I can say
    |x| - |y| >= 0

    But this isn't true as you could have, for example, x = 5 and y = 7. Am I missing something here?
     
  5. Oct 2, 2011 #4

    PeterO

    User Avatar
    Homework Helper

    You were going fine.

    Look at your "which is" line

    You have (....)2 >= 0 I didn't fill in the bracket on purpose.

    Tell me about the value of the square of anything involving real numbers.
     
  6. Oct 2, 2011 #5
    All I can think of is that the square of a real number will be a positive finite real number..?

    So it doesnt matter what is in the brackets it will always be greater than or equal to zero and therefore (|x|-|y|)^2 >=0 or basically (......)^2 >= 0 is true. Is that correct?
     
    Last edited: Oct 2, 2011
  7. Oct 2, 2011 #6
    I am having trouble applying this 'hint' to the original question. The original question was -
    So I have shown that the hint is true but I cant see how I can just substitute in the series [tex]a_n[/tex] and [tex]b_n[/tex] for x and y as the |xy| cannot be represented by [tex]\sum_{n=1}^{\infty}a_nb_n[/tex] as that would be implying that
    [tex]\sum_{n=1}^{\infty}a_n[/tex]*[tex]\sum_{n=1}^{\infty}b_n[/tex] = [tex]\sum_{n=1}^{\infty}a_nb_n[/tex], which it isn't as multiplication of series doesn't work like that. Anyone know how to apply this hint to the original question?
     
    Last edited: Oct 2, 2011
  8. Oct 2, 2011 #7

    PeterO

    User Avatar
    Homework Helper

    That is the logic. The technique of collecting all terms on one side of an equation/inequation is a common approach.
     
  9. Oct 2, 2011 #8
    Cheers for the help mate. Any idea how to apply this hint to the original question in post #6?
     
  10. Oct 3, 2011 #9

    verty

    User Avatar
    Homework Helper

    That hint is really all you need. Perhaps think about what relevance it has? (I'm being terse on purpose, you should know enough to answer this)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook