# Prove: $\mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid$ [duplicate]

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

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?

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

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