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I found this inequality but I don't know where it comes from

  1. Oct 2, 2013 #1
    here is the inequality:
    ##(\sum\limits_{i=1}^n |x_i-y_i|)^2= \ge \sum\limits_{i=1}^n(x_i-y_i)^2+2\sum\limits_{i \neq j}^n |x_i-y_i|\cdot |x_j-y_j|##
    does it have a name/is the consequence of a theorem?
    Thank you :)
  2. jcsd
  3. Oct 2, 2013 #2


    Staff: Mentor

  4. Oct 2, 2013 #3


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    I believe it is actually an equality, assuming you mean the following:
    \left(\sum_{i=1}^{n}|x_i - y_i|\right)^2
    &= \sum_{i=1}^{n} \sum_{j=1}^{n} |x_i - y_i|\cdot |x_j - y_j| \\
    &= \sum_{i=1}^{n} |x_i - y_i|^2 +
    \sum_{i=1}^{n} \sum_{j\neq i,j=1}^{n} |x_i - y_i|\cdot |x_j - y_j|\\
    &= \sum_{i=1}^{n} |x_i - y_i|^2 +
    \sum_{i=1}^{n} \sum_{j=1}^{i-1} |x_i - y_i|\cdot |x_j - y_j| +
    \sum_{i=1}^{n} \sum_{j=i+1}^{n} |x_i - y_i|\cdot |x_j - y_j|
    By symmetry, the second two terms are equal to each other, so we have
    $$\left(\sum_{i=1}^{n}|x_i - y_i|\right)^2 = \sum_{i=1}^{n} |x_i - y_i|^2 + 2\sum_{i=1}^{n} \sum_{j=1}^{i-1} |x_i - y_i|\cdot |x_j - y_j|$$
    I don't know if this equality has a name, but it's a standard and often useful way to rewrite the left hand side.
  5. Oct 2, 2013 #4


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    its the triangle inequality. see the wiki article on it.
  6. Oct 2, 2013 #5

    D H

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    Look at jbunniii's derivation, jedishrfu. It's an identity, not an inequality.
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