Answer
$|\sum\limits_{i =1}^{n}a_{i}| \leq \sum\limits_{i =1}^{n}|a_{i}|$
Work Step by Step
$|\sum\limits_{i =1}^{n}a_{i}| \leq \sum\limits_{i =1}^{n}|a_{i}|$
Use the triangular inequality $|a+b|\leq |a|+|b|$
Expand the summation on both sides.
$a_{1}+a_{2}+....+a_{n}=a_{1}+a_{2}+....+a_{n}$
Take the absolute values.
$|a_{1}+a_{2}+....+a_{n}|=|a_{1}+a_{2}+....+a_{n}|$
Thus,
$|a_{1}+a_{2}+....+a_{n}|\leq|a_{1}|+|a_{2}|+....+|a_{n}|$
(Triangular inequality $|a+b|\leq |a|+|b|$)
Hence, $|\sum\limits_{i =1}^{n}a_{i}| \leq \sum\limits_{i =1}^{n}|a_{i}|$
