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Please review my proof of Cauchy inequality

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm actually only concerned here with proving equality. I would like some review of my proof before I crawl back to my professor again with what I think is a valid proof.

    3. The attempt at a solution
    Show:
    [tex] \frac{x_1+x_2+...+x_n}{n}=\sqrt[n]{x_1x_2\cdots x_n} \Leftrightarrow x_1=x_2=\dots=x_n [/tex]

    Given:
    [tex] \frac{x_1+x_2}{2}=\sqrt{x_1x_2} \Leftrightarrow x_1=x_2 [/tex]

    Assume it is true for [itex] n=k [/itex]
    That is, assume:

    [tex] \frac{x_1+x_2+...+x_k}{k}=\sqrt[k]{x_1x_2\cdots x_k} \Leftrightarrow x_1=x_2=\dots=x_k [/tex]

    for [itex] n=k+1 [/itex] we have:
    [tex] \frac{x_1+x_2+...+x_k+x_{k+1}}{k+1}=\sqrt[k+1]{x_1x_2\cdots x_kx_{k+1}} [/tex]
    Because of the assumption for [itex]k[/itex], we can write:
    [tex] \frac{kx_k+x_{k+1}}{k+1}=\sqrt[k+1]{x_k^kx_{k+1}} [/tex]
    let [itex] x_k-x_{k+1}=\delta [/itex]
    now we can replace [itex]x_k[/itex] by [itex] (x_{k+1}+\delta) [/itex] on one side and [itex] x_{k+1} [/itex] by [itex] (x_k-\delta) [/itex] on the other:
    [tex] \frac{kx_k+x_k-\delta}{k+1}=\sqrt[k+1]{(x_{k+1}+\delta)^kx_{k+1}} [/tex]

    [tex] \lim_{\delta \rightarrow 0} \frac{kx_k+x_k-\delta}{k+1}=\lim_{\delta \rightarrow 0} \sqrt[k+1]{(x_{k+1}+\delta)^kx_{k+1}} [/tex]
    [tex]
    \frac{(k+1)x_k}{k+1}=\sqrt[k+1]{(x_{k+1})^kx_{k+1}}
    [/tex]
    [tex] x_k=x_{k+1} [/tex]
    Thus, equality holds iff [itex] x_1=x_2=\dots =x_n=x_{n+1} [/itex]
    By the Principle of Mathematical Induction, the proof is over.
     
    Last edited: Jul 30, 2011
  2. jcsd
  3. Jul 30, 2011 #2

    I like Serena

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    Hi ArcanaNoir! :smile:

    I guess I can poke a couple of holes in it if you want me to.

    When you take the limit of delta to 0, what you're actually saying is that delta is 0.
    But that is what you have to prove!
    So you've set up a circular proof.

    Btw, the fact that you didn't use the inductive step is a dead give away.

    Furthermore you did not proof the base case, but just assumed it was given.
     
  4. Jul 31, 2011 #3
    Oh bother, I see what you mean.

    what do you mean here?

    I've proven the base case a hundred times this week. I suppose I should prove it the same way I prove the k+1 case though, right? Right.

    Thanks for pointing out the circular logic! :)
     
  5. Jul 31, 2011 #4

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    The inductive step contains a k-root that you assume, and which you would have to use to proof the expression with the (k+1)-root.
    That's what full induction is about.
    As it is you don't use the k-root.
     
  6. Jul 31, 2011 #5
    *sigh* another epic fail. Cauchy=4 me=0.5
     
  7. Jul 31, 2011 #6

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    I'm afraid I have another hole for you, if you're still interested.

    You use the induction assumption for the k-case to conclude that the xi are equal, and then you substitute that in the (k+1) case. But you can't do that, because you don't know that the xi in the (k+1) case are the same as in the k-case.
    Sorry.
     
  8. Jul 31, 2011 #7
    Yeah, I reverted back to my other proof already. This one's pointless. I should have known it was nonsense when it came out all concise and visually appealing.
     
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