# Is this a valid proof of the triangle inequality?

Inequality to prove:

$|a+b|\leq |a| + |b|$

Proof:

1. $-|a| \leq a \leq |a|$

2. $-|b| \leq b \leq |b|$

Add 1 and 2 together to get:

$-(|a|+|b|)\leq a+b\leq|a|+|b|$

$|a+b|\leq|a|+|b|$

The reason I'm asking is because this looks like the simplest proof of all proofs I've seen but it's rarely used. I am wondering why more "complicated" proofs are being used. Is there something wrong with this proof?

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The complicated proofs are for $a,b\in R^n$. For $n=1$ everything is simple. – Fabian Jan 5 '13 at 13:01
Well, the second-to-last line of mathematics in your question actually means that $$|a+b|\leq \left|\,|a|+|b|\,\right|\ldots$$ This is a tad short of what you actually want, but you've made almost all. – DonAntonio Jan 5 '13 at 13:02
Yours is perfect. Now replace $a$ by $a-b$ in the last inequality to find $|a-b|\geq|a|-|b|$ – Babak S. Jan 5 '13 at 13:02
Note that $-y\le x\le y$ implies $x\le y\land -x\le y$ and hence $|x|=\max\{x,-x\}\le y$. – Hagen von Eitzen Jan 5 '13 at 13:03
This proof is hard to generalize by another cases, For example, $\mathbb{R}^n$ and $\mathbb{C}$. – Hanul Jeon Jan 5 '13 at 13:11

The proof is indeed simple for $x, y \in \mathbb{R}$. Things get more complicated for $\mathbb{R}^n$ where you cannot compare $x$ with $|x|$. This is also true for metric spaces in general.