# Proof of triangle inequality

I understand intuitively that this is true, but I'm embarrassed to say I'm having a hard time constructing a rigorous proof that $|a+b| \leq |a|+|b|$. Any help would be appreciated :)

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Isn't this an axiom in metric space? –  Shu Xiao Li Feb 18 '13 at 19:11
There is no addition in metric space. @ShuXiaoLi –  k.stm Feb 18 '13 at 19:16

Prove $|x| = \max\{x,-x\}$ and $\pm x ≤ |x|$.

Then you can use: \begin{align*} a + b &≤ |a| + b ≤ |a| + |b|,\quad\text{and}\\ -a - b &≤ |a| -b ≤ |a| + |b|. \end{align*}

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Clear and concise, +1. –  1015 Feb 18 '13 at 19:18
Nice. Thanks! I got hung up for a sec on this step:$$-a-b\leq|a|-b$$ but then I put in the intermediate step:$$-a-b\leq|-a|-b=|a|-b$$Thank you! –  ivan Feb 18 '13 at 19:28

$$a^2+b^2+2|a||b|\geq a^2+b^2+2ab$$ $$(|a|+|b|)^2 \geq |a+b|^2\phantom{a}(\because \forall x\in \mathbb{R};\phantom{;}x^2=|x|^2)$$ $$\therefore |a|+|b|\geq |a+b|$$

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If a neat algebraic argument does not suggest itself, we can do a crude argument by cases, guided by the examples $a=7,b=4$, $a=-7,b=-4$, $a=7, b=-4$, and $a=-7, b=4$.

If $a\ge 0$ and $b\ge 0$ then $|a+b|=|a|+|b|$.

If $a\le 0$, and $b\le 0$, then $|a+b|=-(a+b)=(-a)+(-b)=|a|+|b|$.

Now we need to examine the cases where $a$ is positive and $b$ is negative, or the other way around. Without loss of generality we may assume that $|b|\le |a|$.

If $a\gt 0$, then $|a+b|=|a|-|b|$. This is $\lt |a|$, and in particular $\lt |a|+|b|$.

If $a\lt 0$, then again $|a+b|=|a|-|b|$.

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