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Math Help - Minkovski's Inequality

  1. #1
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    Minkovski's Inequality

    We were given this inequality to prove:

    sqrt(summation of (ak + bk)^2 from K=1 to n) <=

    sqrt(summation of (ak)^2 from K=1 to n) + sqrt(summation of (bk)^2 from K=1 to n)

    We were told to square both sides and then use the Cauchy-Schwarz inequality to finish.
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  2. #2
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    Quote Originally Posted by Judi
    We were given this inequality to prove:

    sqrt(summation of (ak + bk)^2 from K=1 to n) <=

    sqrt(summation of (ak)^2 from K=1 to n) + sqrt(summation of (bk)^2 from K=1 to n)

    We were told to square both sides and then use the Cauchy-Schwarz inequality to finish.
    You need to show,
    0\leq \sqrt{ \sum_{k=1}^n (ak+bk)^2 }\leq \sqrt{ \sum_{k=1}^n(ak)^2} +\sqrt{ \sum_{k=1}^n(bk)^2 }
    Upon squaring we have, if and only if
    \sum_{k=1}^n (ak+bk)^2\leq \sum_{k=1}^n(ak)^2+\sum_{k=1}^n(bk)^2+2\sqrt{\sum_  {k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2
    If and only If
    \sum_{k=1}^n (ak)^2+2(ak)(bk)+(bk)^2\leq  \sum_{k=1}^n(ak)^2+\sum_{k=1}^n(bk)^2+2\sqrt{\sum_  {k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2
    if and only if
    2\sum_{k=1} (ak)(bk)\leq 2\sqrt{\sum_{k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2
    if and only if
    \sum_{k=1}^n (ak)(bk) \leq \sqrt{\sum_{k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2
    If and only if *)
    \left( \sum_{k=1}^n (ak)(bk) \right)^2 \leq \sum_{k=1}^n (ak)^2\cdot \sum_{k=1}^n (bk)^2
    This is true by Cauchy-Swarchz Inequality.
    ----
    Explanation of last step.
    We will prove a famous case,
    \left( \sum_{k=1}^n a_k b_k\right)^2\leq \left(\sum_{k=1}^n a_k^2 \right) \cdot \left( \sum_{k=1}^n b_k^2 \right)
    Because this is true only when, (take square roots)
    \left| \sum_{k=1}^n a_k b_k \right| \leq \sqrt{\sum_{k=1}^n a_k^2}\cdot \sqrt{\sum_{k=1}^n b_k^2 \right)

    Consider vectors in \mathbb{R}^n
    \bold{v}=(a_1,a_2,...,a_n)
    \bold{u}=(b_1,b_2,...,b_n)
    Then,
    |\bold{v} \cdot \bold{u}|\leq ||\bold{v}||\cdot ||\bold{u}||-by Cauchy-Swartchz
    But,
    |\bold{v}\cdot \bold{u}|=\left| \sum_{k=1}^n a_k b_k \right|
    and,
    ||\bold{v}||\cdot ||\bold{u}||=\sqrt{\sum_{k=1}^n a_k^2}\cdot \sqrt{\sum_{k=1}^n b_k^2 \right)
    \mathbb{Q}.\mathbb{E}.\mathbb{D}



    *)If LHS is negative there is nothing to prove because RHS must be positive! We assume it is positive also.

    Note: The consistent use of 'if and only if' otherwise the proof falls.
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