#### Answer

See proof below.

#### Work Step by Step

1. (base) Check that the formula holds for n = 1.
LHS$=|x_{1}|,\qquad $RHS$=|x_{1}|$
$|x_{1}|\leq|x_{1}|\qquad $... yes, it is true when $n=1.$
2. (step) Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$.
Let us assume that for $n=k$,
$|x_{1}+x_{2}+...+x_{k}|\leq|x_{1}|+|x_{2}|+...+|x_{k}|$
(we call this the step hypothesis)
For $n=k+1$, the LHS equals $|x_{1}+x_{2}+...+x_{k}+x_{k+1}|$
If we set $a=x_{1}+x_{2}+...+x_{k}$ and $b=x_{k+1}$,
the LHS can be rewritten as $|a+b|$.
We now apply the triangle inequality
$|a+b|\leq|a|+|b|$
$|x_{1}+x_{2}+...+x_{k}+x_{k+1}|\leq|x_{1}+x_{2}+...+x_{k}|+|x_{k+1}|$
By the step hypothesis, $|x_{1}+x_{2}+...+x_{k}|\leq|x_{1}|+|x_{2}|+...+|x_{k}|$
so we can say that
$|x_{1}+x_{2}+...+x_{k}+x_{k+1}|\leq|x_{1}|+|x_{2}|+...+|x_{k}|+|x_{k+1}|,$
which makes the formula true for $n=k+1.$
By mathematical induction, the formula is true for any integer $n.$