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This question already has an answer here:

$\mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid$

If $n$ is even we will divide the sum into groups of $2$ $x$'s namely $\mid x+x \mid \leq \mid x\mid+\mid x \mid$ and will repeat the process to get $\mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid$

If $n$ is odd, we will divide the sum into an even number of $x$'s called $a$ and the leftover $x$ called $b$, by using the proof for even $n$ and the triangle inequality

$\mid a+b\mid\leq \mid a \mid +\mid b \mid= \mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid$

Is the proof valid?

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marked as duplicate by Dietrich Burde, Community Jul 6 '16 at 9:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Use the triangle inequality (this is the case when $n=2$): $|x_{1}+x_{2}|\leq |x_{1}|+|x_{2}|$. Then, induct on $n$. $\endgroup$ – Karthik Jul 6 '16 at 9:30
  • $\begingroup$ Use $\pm x_i\le\mid x_i\mid$ so $\pm\sum x_i\le\sum\mid x_i\mid.$ $\endgroup$ – awllower Jul 6 '16 at 9:38