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Math Help - 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 n\le k)
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  3. #3
    Senior Member
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    write the inductive hypothesis as
    \left(  \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{  2}}+\cdots<br />
+\sqrt{a_{k}^{2}+b_{k}^{2}}\right)<br />
^{2} \geq \left(  \sum\limits_{i=1}^{k}a_{i}^{2}\right)<br />
^{2}+\left(  \sum\limits_{i=1}^{k}b_{i}^{2}\right)<br />
^{2}
    Let RHS denote the right hand side of the above inequality.
    Notice that
    \left(  \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{  2}}+\cdots<br />
+\sqrt{a_{k}^{2}+b_{k}^{2}}+\sqrt{a_{k+1}^{2}+b_{k  +1}^{2}}\right)<br />
^{2} =\left(  \sum\limits_{i=1}^{k}\sqrt{a_{i}^{2}+b_{i}^{2}}\ri  ght)<br />
^{2}+2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left(  \sum\limits_{i=1}^{k}\sqrt<br />
{a_{i}^{2}+b_{i}^{2}}\right)  +\left(  \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right)<br />
^{2}

    \left(  \sum\limits_{i=1}^{k+1}a_{i}\right)  ^{2} =\left(  \sum\limits_{i=1}%<br />
^{k}a_{i}\right)  ^{2}+ 2a_{k+1}\left(  \sum\limits_{i=1}^{k}a_{i}\right)<br />
+a_{k+1}^{2}

    \left(  \sum\limits_{i=1}^{k+1}b_{i}\right)  ^{2} =\left(  \sum\limits_{i=1}%<br />
^{k}b_{i}\right)  ^{2}+ 2b_{k+1}\left(  \sum\limits_{i=1}^{k}b_{i}\right)<br />
+b_{k+1}^{2}

    From the inductive hypothesis,

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

    and now it only remain to show that

    2\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\left(  \sum\limits_{i=1}^{k}\sqrt{a_{i}%<br />
^{2}+b_{i}^{2}}\right)  +\left(  \sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right)<br />
^{2} \geq2a_{k+1}\left(  \sum\limits_{i=1}^{k}a_{i}\right)  +a_{k+1}%<br />
^{2}+2b_{k+1}\left(  \sum\limits_{i=1}^{k}b_{i}\right)  +b_{k+1}^{2}

    Hence,

    \left(  \sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{  2}}+\cdots<br />
+\sqrt{a_{k+1}^{2}+b_{k+1}^{2}}\right)<br />
^{2} \geq \left(  \sum\limits_{i=1}^{k+1}a_{i}^{2}\right)<br />
^{2}+\left(  \sum\limits_{i=1}^{k+1}b_{i}^{2}\right)<br />
^{2}
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