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

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Felafel
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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 :)
 
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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.
 
its the triangle inequality. see the wiki article on it.
 
Look at jbunniii's derivation, jedishrfu. It's an identity, not an inequality.
 
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