# 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? – NECing Feb 18 '13 at 19:11
There is no addition in metric space. @ShuXiaoLi – k.stm Feb 18 '13 at 19:16
That a metric must obey the triangle inequality is indeed one of the axioms of a metric space. – Mathematacticool Jul 28 '15 at 1:04

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
Do you need that step though? Because $|x|=max\{x,-x\}$, which is trivially greater than or equal to $-x$. – sodiumnitrate Mar 8 '15 at 3:32

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

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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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A simple proof of the triangle inequality that is complete and easy to understand (there are more cases than strictly necessary; however, my goal is clarity, not conciseness).

Prove the triangle inequality $| x | + | y| ≥ | x + y|$.

Without loss of generality, we need only consider the following cases:

1. $x = 0$
2. $x > 0, y > 0$
3. $x > 0, y < 0$

Case $1$. Suppose $x = 0$. Then we have

$| x| = 0$

$| x| + | y| = 0 + | y| = | y|$

Thus $| x| + | y| = | x + y|$.

Case $2$. Suppose $x > 0, y > 0$. Then, since $x + y > 0$, we have

$| x| = x$

$| y| = y$

$| x| + | y| = x + y$

$| x + y| = x + y$

Thus $| x| + | y| = | x + y|$.

Case $3$. Suppose $x < 0, y < 0$. Then, since $x + y < 0$, we have

$| x| = −x$

$| y| = −y$

$| x| + | y| = (−x) + (−y)$

$| x + y| = −(x + y) = (−x) + (−y)$

Thus $| x| + | y| = | x + y|$.

Case $4$. Suppose $x > 0, y < 0$. Then we have

$| x| = x$

$| y| = −y$

$| x| + | y| = x + (−y)$

We must now consider three cases:

a. $x + y = 0$

b. $x + y > 0$

c. $x + y < 0$

Case $4a$. Suppose $x + y = 0$. Then we have

$| x + y | = |0| = 0$

Since $y < 0$, it follows that $−y > 0$ and thus $x + (−y) > 0 + (-y) = -y > 0$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

Case $4b$. Suppose $x + y > 0$. Then we have

$| x + y| = x + y$

Since $y < 0$, it follows that $−y > 0 > y$ and thus $x + (−y) > x + y$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

Case $4c$. Suppose $x + y < 0$. Then we have

$| x + y| = −(x + y) = (−x) + (−y)$

Since $x > 0$, it follows that $x > 0 > −x$ and thus $x + (−y) > (−x) + (−y)$.

Therefore, since $| x| + | y| = x + (−y)$, we must have $| x| + | y| > | x + y|$.

This concludes the proof.

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