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

  • Thread starter Felafel
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  • #1
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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 :)
 

Answers and Replies

  • #3
jbunniii
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I believe it is actually an equality, assuming you mean the following:
$$\begin{align}
\left(\sum_{i=1}^{n}|x_i - y_i|\right)^2
&= \sum_{i=1}^{n} \sum_{j=1}^{n} |x_i - y_i|\cdot |x_j - y_j| \\
&= \sum_{i=1}^{n} |x_i - y_i|^2 +
\sum_{i=1}^{n} \sum_{j\neq i,j=1}^{n} |x_i - y_i|\cdot |x_j - y_j|\\
&= \sum_{i=1}^{n} |x_i - y_i|^2 +
\sum_{i=1}^{n} \sum_{j=1}^{i-1} |x_i - y_i|\cdot |x_j - y_j| +
\sum_{i=1}^{n} \sum_{j=i+1}^{n} |x_i - y_i|\cdot |x_j - y_j|
\end{align}$$
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.
 
  • #4
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its the triangle inequality. see the wiki article on it.
 
  • #5
D H
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Look at jbunniii's derivation, jedishrfu. It's an identity, not an inequality.
 

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