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Cauchy–Schwarz inequality

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove that:

    2. Relevant equations

    [tex]\sum_{k=1}^{n}x_{k}^{2}\geq \frac{1}{n}\left ( \sum_{k=1}^{n}x_{k} \right )^{2}[/tex]

    3. The attempt at a solution

    I am not sure what to do to be honest. But it looks like the Cauchy–Schwarz inequality to me.
     
  2. jcsd
  3. Feb 16, 2013 #2

    micromass

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    Re: Proof

    Oh come on. I'm sure there are things you can try. For example, solve the case n=2. Or try induction. Or something.
     
  4. Feb 16, 2013 #3
    Re: Proof

    Sigma is intimidating :(
    But I will try doing it.
     
  5. Feb 16, 2013 #4

    micromass

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    Re: Proof

    Yes, I agree that it is intimidating. That is why I usually do some special cases first like n=2 and n=3 to get a feeling for the proof!
     
  6. Feb 16, 2013 #5
    Re: Proof

    I got [tex]2x_{1}^{2}+2x_{2}^{2}\geq x_{1}^{2}+2x_{1}x_{2}+x_{2}^{2}[/tex]
     
  7. Feb 16, 2013 #6

    micromass

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    Re: Proof

    Do you see why this is true?
     
  8. Feb 16, 2013 #7
    Re: Proof

    Is it because [tex]\left ( x_{1}^{2}+x_{2}^{2} \right )\geq 2x_{1}x_{2}[/tex]
    But why is this true?
    PS: Is it because [tex]\left ( x_{1}-x_{2} \right )^{2}\geq 0 \therefore \left ( x_{1}^{2}+x_{2}^{2} \right )\geq 2x_{1}x_{2}[/tex]
     
  9. Feb 16, 2013 #8
    Re: Proof

    Do I just do normal induction from here?
     
  10. Feb 16, 2013 #9

    micromass

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    Re: Proof

    After thinking a little bit about it, you won't need induction at all.

    Basically, you need to prove that

    [tex]\sum_{k=1}^n x_k \leq \sqrt{n}\sqrt{\sum_{k=1}^n x_k^2}[/tex]

    You were completely right that this resembled Cauchy-Schwarz. So, try to apply Cauchy-Schwarz somewhere.

    Of course, for that inequality, you need two sequences. The sequence [itex](x_1,...,x_n)[/itex] is one of them, what is the other?
     
  11. Feb 16, 2013 #10
    Re: Proof

    Is it [tex]x_{1}^{2},...,x_{n}^{2}[/tex] ?

    And thank you a lot for the help. I didn't think about squarerooting both sides :( I really need to learn to recognized this stuff.
     
  12. Feb 16, 2013 #11

    micromass

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    Re: Proof

    No, it's not that.

    Think of a very easy (constant) sequence.
     
  13. Feb 17, 2013 #12
    Re: Proof

    Although this question can be done using the CS Inequality, it can be done this way too:
    [tex]a^2+b^2 \geq 2ab [/tex]

    [We need to sub into a and b, all the combinations of 2 numbers from the set [tex]\ \ x_1, \ x_2, \ x_2 \ \dots, \ x_n [/tex]

    [tex]x_1^2+x_1^2 \geq 2x_1 x_2 [/tex]

    [tex]x_1^2+x_3^2 \geq 2x_1x_3 [/tex]

    .
    .
    .

    [tex]x_1^2+x_n^2 \geq 2x_1x_n [/tex]

    [tex]x_2^2+x_3^2 \geq 2x_2 x_3 [/tex]

    And so on, then we add these inequalities side by side.

    [tex](n-1)(x_1^2+x_2^2 \dots + x_n^2) \geq 2(x_1x_2+x_1x_3+ \dots) [/tex]

    [tex]n(x_1^2+x_2^2+\dots+x_n^2) \geq x_1^2+x_2^2+\dots + x_n^2 + 2(x_1x_2+\dots [/tex]

    The RHS is the expanded form of:

    [tex]n\sum_{k=1}^{n} x_k^2 \geq \left(\sum_{k=1}^{n}x_k \right)^2 [/tex]

    [tex]\sum_{k=1}^{n} x_k^2 \geq \frac{1}{n}\left(\sum_{k=1}^{n}x_k \right)^2 [/tex]
     
  14. Feb 17, 2013 #13
    Re: Proof

    Could you please show me how you would do it using the CS inequality?
     
  15. Feb 17, 2013 #14

    micromass

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    Re: Proof

    You need to show that


    [tex]\sum_{k=1}^n x_k \leq \sqrt{n}\sqrt{\sum_{k=1}^n x_k^2}[/tex]

    The CS inequality gives something like

    [tex]\sum_{k=1}^n x_ky_k \leq ...[/tex]

    If I use the same [itex]x_k[/itex] in both inequalities above, then what does the corresponding [itex]y_k[/itex] need to be?
     
  16. Feb 17, 2013 #15
    Re: Proof

    Ok, thank you. I will try to do it now.
     
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