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When does equality hold? schwarz inequality

  1. Sep 7, 2016 #1
    1. The problem statement, all variables and given/known data
    By choosing the correct vector b in the Schwarz inequality, prove that (a1 + ... + an)^2 =< n(a1^2 + ... +an^2)

    2. Relevant equations
    Schwarz inequality

    3. The attempt at a solution
    since the answer key says that a1 = a2 = ... = an, i tried plugging in values, but i am not getting anywhere and i completely have no intuition of what is happening here. and while on the topic, can anyone recommend books to read - both introductory and more advanced ones about inequalities? i would like a book that gives visual and geometric intuitions. books with proofs would be nice, but it's not a priority to me just yet. thank you!

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  2. jcsd
  3. Sep 7, 2016 #2


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    Gold Member

    You can easily check that for ##a_1 = a_2 = \cdots = a_n## the equality holds. As a hint, use the Cauchy - Schwarz inequality with ##b = (1,1,1,..,1)##. You need one more vector a. I leave it to you to figure it out and apply the inequality.

    EDIT: I would recommend the book Inequalities by Hardy - Littlewood - Polya https://www.amazon.com/Inequalities-Cambridge-Mathematical-Library-Hardy/dp/0521358809. This classic, is a comprehensive study of inequalities. For introductory level, I recommend practicing over a lot of exercises, that can be found easily on the net. But of course, there are lots of good introductory texts too.
    Last edited by a moderator: May 8, 2017
  4. Sep 7, 2016 #3
    thanks i think i got it but i don't get what the "n" in front of "n(a1^2 + ... +an^2)" means
    Last edited by a moderator: May 8, 2017
  5. Sep 7, 2016 #4
    is n = (<1,1,...,1>)^2...?
  6. Sep 7, 2016 #5


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    In order to understanding it better, give some small value to n and see what the ##(a_1 + \cdots + a_n)^{2} \leq n(a_1^{2}+\cdots+a_n^{2})## gives.
  7. Sep 7, 2016 #6
    ahh got it! thanks. that's so simple that it's embarassing
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