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 :)
From your definition of the absolute value, establish first $|x| = \max\{x,-x\}$ 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*}
$$a^2+b^2+2|a||b|\geq a^2+b^2+2ab$$ $$(|a|+|b|)^2 \geq |a+b|^2\phantom{a}(\because \forall x\in \mathbb{R};\phantom{;}x^2=|x|^2)$$ $$\therefore |a|+|b|\geq |a+b|$$
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:
- $x = 0$
- $x > 0, y > 0$
- $x > 0, y < 0$
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.
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 $|a+b|=|a|-|b|$.
The proof given in Wikipedia / Absolute Value is interesting and the technique can be used for complex numbers:
Choose $\epsilon$ from $\{ -1,1\}$ so that $\epsilon (a+b) \ge 0$. Clearly $\epsilon x \le |x|$ for all real $x$ regardless of the value of $\epsilon$, so
$|a+b|= \epsilon (a+b) = \epsilon a + \epsilon b \le |a|+|b|$