- #1

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##(\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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- Thread starter Felafel
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- #1

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##(\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

jedishrfu

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- #3

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$$\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

jedishrfu

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its the triangle inequality. see the wiki article on it.

- #5

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

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