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Cauchy -schwarz inequality help

  1. Jul 5, 2010 #1
    Need help proving Cauchy Schwarz inequality ...

    the first method I know is pretty easy

    [itex]\displaystyle\sum_{i=1}^n (a_ix-b_i)^2 \geq 0 [/itex]

    expanding this and using the discriminatant quickly establishes the inequality..

    The 2nd method I know is I think a easier one , but I dont have a clue about how this notation works..

    Since cauchy SHwarz inquality states..

    [tex](a_1b_1+a_2b_2+...+a_nb_n)^2 \leq ((a_1)^2+(a_2)^2+..+(a_n)^2)((b_1)^2+(b_2)^2+...+(b_n)^2)[/tex]

    [tex]((a_1)^2+(a_2)^2+..+(a_n)^2)((b_1)^2+(b_2)^2+...+(b_n)^2)-(a_1b_1+a_2b_2+...+a_nb_n)^2 \geq 0[/tex]

    I dont usnderstand how the below notation works as I cant follow from the above line to the line below , if someone can point me to some resources where I can know more about it :) ...

    [itex]\displaystyle\sum_{i\not=j}^n ((a_i)^2(b_j)^2+(a_j)^2(b_i)^2-2a_ib_ja_jb_i ) [/itex]

    Last edited: Jul 5, 2010
  2. jcsd
  3. Jul 5, 2010 #2
    That's just summing over the quantities within the parentheses for which the two indices i and j differ. Note that if i = j, the quantity inside the summation is just 0, so it does not contribute to the sum.
  4. Jul 5, 2010 #3
    thanks :) , one more small question when they have summed the above where is [tex]-2a_ib_ja_jb_i[/tex] comming from?
  5. Jul 5, 2010 #4
    [tex](a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2[/tex]​

    Expanded out, convince yourself that whenever i ≠ j you get a 2aibiajbj. To do this, it helps to write out a simple case for small n (n = 2 or 3) and look at how what terms you get that involve i ≠ j. The sign in your example is negative, because you are subtracting.
  6. Jul 5, 2010 #5
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