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Thread: Proof by induction stuck at very end!!

  1. #1
    Senior Member
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    Proof by induction stuck at very end!!

    Show that for all real values of the variables. Furthermore, give a condition for equality to hold.

    Base case:

    Let











    Which is true because of Cauchy-Schwarz.

    Inductive Hypothesis:

    Assume the inequality is true for



    is true.

    Proof:

    Need to prove it's true for



    So let's add to our inductive hypothesis.

    This yields:



    now what?
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  2. #2
    Senior Member Shanks's Avatar
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    It would be better to prove it By using the Second Mathematical Induction(change the inductive hypothesis to $\displaystyle n\le k$)
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  3. #3
    Senior Member
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    write the inductive hypothesis as
    $\displaystyle \left( \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{ 2}}+\cdots
    +\sqrt{a_{k}^{2}+b_{k}^{2}}\right)
    ^{2}$$\displaystyle \geq \left( \sum\limits_{i=1}^{k}a_{i}^{2}\right)
    ^{2}+\left( \sum\limits_{i=1}^{k}b_{i}^{2}\right)
    ^{2}$
    Let$\displaystyle RHS$ denote the right hand side of the above inequality.
    Notice that
    $\displaystyle \left( \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{ 2}}+\cdots
    +\sqrt{a_{k}^{2}+b_{k}^{2}}+\sqrt{a_{k+1}^{2}+b_{k +1}^{2}}\right)
    ^{2}$$\displaystyle =\left( \sum\limits_{i=1}^{k}\sqrt{a_{i}^{2}+b_{i}^{2}}\ri ght)
    ^{2}+2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}\sqrt
    {a_{i}^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right)
    ^{2}$

    $\displaystyle \left( \sum\limits_{i=1}^{k+1}a_{i}\right) ^{2}$$\displaystyle =\left( \sum\limits_{i=1}%
    ^{k}a_{i}\right) ^{2}+$$\displaystyle 2a_{k+1}\left( \sum\limits_{i=1}^{k}a_{i}\right)
    +a_{k+1}^{2}$

    $\displaystyle \left( \sum\limits_{i=1}^{k+1}b_{i}\right) ^{2}$$\displaystyle =\left( \sum\limits_{i=1}%
    ^{k}b_{i}\right) ^{2}+$$\displaystyle 2b_{k+1}\left( \sum\limits_{i=1}^{k}b_{i}\right)
    +b_{k+1}^{2}$

    From the inductive hypothesis,

    $\displaystyle \left( \sum\limits_{i=1}^{k}\sqrt{a_{i}^{2}+b_{i}^{2}}\ri ght) ^{2}%
    +2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}\sqrt{a_{i}%
    ^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right)
    ^{2}$$\displaystyle \geq RHS+2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}%
    \sqrt{a_{i}^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}%
    }\right) ^{2}$

    and now it only remain to show that

    $\displaystyle 2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left( \sum\limits_{i=1}^{k}\sqrt{a_{i}%
    ^{2}+b_{i}^{2}}\right) +\left( \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right)
    ^{2}$$\displaystyle \geq2a_{k+1}\left( \sum\limits_{i=1}^{k}a_{i}\right) +a_{k+1}%
    ^{2}+2b_{k+1}\left( \sum\limits_{i=1}^{k}b_{i}\right) $$\displaystyle +b_{k+1}^{2}$

    Hence,

    $\displaystyle \left( \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{ 2}}+\cdots
    +\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right)
    ^{2}$$\displaystyle \geq \left( \sum\limits_{i=1}^{k+1}a_{i}^{2}\right)
    ^{2}+\left( \sum\limits_{i=1}^{k+1}b_{i}^{2}\right)
    ^{2}$
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