Proving by induction that $| x_1 + x_2 + ... + x_n | leq | x_1 | + | x_2| + ... + | x_n |$

This is the first question in Thomas' Calculus Appendix on Proof by Induction Exercise (Exercise A.1).

As the title suggests, I'd like to prove by induction that $| x_1 + x_2 + ... + x_n | \leq | x_1 | + | x_2| + \ldots + | x_n |$ is true for any n numbers.

You are told to assume that the triangle inequality $|a+b| \leq |a| + |b|$ is true.

I'm 17 and I've only done Proof By Induction in Further Pure 1 (A first year module in Further Pure Mathematics at College, in the UK), so sorry if this seems incredibly simple. So far I have this:

$$
\text{Let n = 2 } \\
\implies LHS = | 1 + 2 | = |3| = 3 \\
\implies RHS = | 1 | + | 2| = 1 + 2 = 3 \\
\text{Therefore I have proved it for n = 2 } \\
\text{Assume that $n=k$ is true, Let $n = k+1$}

$$

And that's about it :).

I know how to say it in words; that the LHS is the absolute value of the sum of all n numbers, therefore when $x \in \mathbb{R^-}$ the actual value of the sum of all n numbers could be negative, where as the RHS is the sum absolute value of each $x$... but I don't know how to prove it...

Thanks.

Answer

In the induction step you have $|x_1+\ldots+x_n|\le |x_1|+\ldots+|x_n|$ and want ot show $|x_1+\ldots+x_n+x_{n+1}|\le |x_1|+\ldots+|x_n|+|x_{n+1}|$.
Simply plug $a=x_1+\ldots+x_n$, $b=x_{n+1}$ into the triangle inequality to obtain

$$\begin{matrix}|x_1+\ldots+x_n+x_{n+1}|&=&|a+b|\le|a|+|b|\\&=& |x_1+\ldots+x_n|+|x_{n+1}|\\&\le& |x_1|+\ldots+|x_n|+|x_{n+1}|.\end{matrix}$$