# 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.

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I don't thikn you are proving the what is being asked. What you need to do is use that for $n=1$ $$|x_1 + x_2| \le |x_1| + |x_2|$$ by the triangle inequality, assume $$|x_1 + \ldots + x_n| \le |x_1| + \ldots + |x_n|$$ by induction hypotesis, and prove $$|x_1 + \ldots + x_{n+1}| \le |x_1| + \ldots + |x_{n+1}|$$ – Pragabhava Nov 11 '12 at 18:46

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 $$\begin{matrix}|x_1+\ldots+x_n+x_{n+1}|&=&|a+b|\le|a|+|b|\\&=& |x_1+\ldots+x_n|+|x_{n+1}|\\&\le& |x_1|+\ldots+|x_n|+|x_{n+1}|.\end{matrix}$$
You seem to be confused at what $n$ is. It is supposed to denote the number of things being summed. These numbers are not necessarily the numbers $1, 2, \dots, n$, otherwise the result is trivial.
Hint. $x_1+\dots + x_n=(x_1+\dots + x_{n-1})+x_n$.