Show it for n=1,2. This may take a few cases and on how you defined absolute value.
IH: |x1+...+xn| < |x1| + ... + |xn| for n<=k.
Let Y = x1+...+xk, and note by our IH, |Y| <= |x1|+...+|xk|. (Y is just a real number now, expressable as this finite sum)
|x1 + ... + xk + x(k+1)| = |Y+x(k+1)| <= |Y| + |x(k+1)| <= |x1|+|x2|+...+|xk| + |x(k+1)|