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Thread: Euclidean metric is a metric

  1. #1
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    Euclidean metric is a metric

    I am trying to prove that the euclidean metric is a metric. I was able to prove the first three properties easily but the fourth is giving me trouble.

    I have to prove that

    $\displaystyle \displaystyle\sqrt{\sum_{i=1}^{n}(x_i-z_i)^2}\leq \sqrt{\sum_{i=1}^{n}(x_i-y_i)^2}+\sqrt{\sum_{i=1}^{n}(y_i-z_i)^2}$

    Now in the outlined proof they ask you to first prove the CBS inequality and obtain the form

    $\displaystyle \displaystyle\sum_{i=1}^{n}x_iy_i \leq \sqrt{\left(\sum_{i=1}^{n}x_i^2\right)\left(\sum_{ i=1}^{n}y_i^2\right)}$

    And I've done this.

    What I don't get is the jump from this to the proof of the fourth condition.

    Could someone give me a hint (I'd actually prefer a hint rather then the whole solution if possible or a guide I want to get as much of this on my own as possible.)

    Also, do I need to use $\displaystyle \sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ and if so how should I approach proving that?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    $\displaystyle d^2(x,z)=\displaystyle\sum_{i=1}^n{(x_i-z_i)^2} =\displaystyle\sum_{i=1}^n(x_i-y_i+y_i-z_i)^2=$

    $\displaystyle \displaystyle\sum_{i=1}^n(x_i-y_i)^2+\displaystyle\sum_{i=1}^n(y_i-z_i)^2+2\displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)$

    Now, use Cauchy Schwartz inequality:

    $\displaystyle \displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)\leq \left|\displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)\right|\leq $

    $\displaystyle \sqrt{\displaystyle\sum_{i=1}^n{(x_i-y_i)^2}}\sqrt{\displaystyle\sum_{i=1}^n{(y_i-z_i)^2}} $

    You'll obtain:

    $\displaystyle d^2(x,z)\leq \left(d(x,y)+d(y,z)\right)^2$


    Fernando Revilla
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  3. #3
    Junior Member
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    $\displaystyle d^2(x,z)=\displaystyle\sum_{i=1}^n{(x_i-z_i)^2} =\displaystyle\sum_{i=1}^n(x_i-y_i+y_i-z_i)^2$

    Makes everything crystal clear.

    So from

    $\displaystyle \displaystyle\sum_{i=1}^n(x_i-y_i)^2+\displaystyle\sum_{i=1}^n(y_i-z_i)^2+2\displaystyle\sum_{i=1}^n(x_i-y_i)(y_i-z_i)$

    We get

    $\displaystyle \displaystyle\sum_{i=1}^n(x_i-y_i)^2+\displaystyle\sum_{i=1}^n(y_i-z_i)^2+2\sqrt{\displaystyle\sum_{i=1}^n{(x_i-y_i)^2}}\sqrt{\displaystyle\sum_{i=1}^n{(y_i-z_i)^2}}$

    Which becomes

    $\displaystyle \displaystyle\sum_{i=1}^n{(x_i-z_i)^2\leq\left(\sqrt{\displaystyle\sum_{i=1}^n(x_ i-y_i)^2}+\sqrt{\displaystyle\sum_{i=1}^n(y_i-z_i)^2}\right)^2$

    $\displaystyle \sqrt{\displaystyle\sum_{i=1}^n{(x_i-z_i)^2}}\leq\sqrt{\displaystyle\sum_{i=1}^n(x_i-y_i)^2}+\sqrt{\displaystyle\sum_{i=1}^n(y_i-z_i)^2}$

    Thank you so much for the help.
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