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From page 11 (4th edition Spivak's calculus).

First off, a definition for $|a|$ would be $$\left| a \right| = \begin{cases} a, \qquad a \geq 0 \\ -a, \qquad a \leq 0 \end{cases}$$

Theorem: For all numbers $a$ and $b$, we have $$|a+b| \leq |a| + |b|$$

and concentrating only on 1 of the 4 cases ie let $a \geq 0$ and $b \geq 0$ we have

$|a+b| = a + b = |a| + |b|$

It's the second and third expression, the assertion that $a + b = |a| + |b|$ that is unclear to me. Clearly $|a+b|$ must be $\geq 0$, and by definition allows the assertion $|a + b| = a + b$.

Perhaps another question might shed some light. Assume that $z \geq 0$. Can I simply thus assert that $ z = |z|$? If this sounds a little confusing, it's probably because I am missing something.

Kyan Cheung
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user1115542
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    By your definition, yes $z\geq 0\iff |z|=z$ (for real $z$ btw) – Kyan Cheung Oct 09 '21 at 14:18
  • It's truly that simple?? Reals, understood :-) – user1115542 Oct 09 '21 at 14:20
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    Absolute value means non-negative value so naturally
    $z \ge 0 \implies |z|=z$     – poetasis Oct 09 '21 at 14:34
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    One could also use the Fundamental theorem of Absolute values, i.e., $-M \leq a \leq M \iff |a| \leq M$. Apply this for two real numbers $a, b$ -- where $M=|a|$ and $M=|b|$ respectively -- and add the inequality to get your desired expression to prove the theorem. This approach I find much nicer than 4 uninteresting cases. – Owen Murphy Oct 11 '21 at 23:07
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    Notice that the definition of $|a+b|$ would be
    $$\left| a+b \right| = \begin{cases} a+b, \qquad a+b \geq 0 \ -(a+b), \qquad a+b \leq 0 \end{cases}$$ by which $$\left| a \right| = \begin{cases} a, \qquad a \geq 0 \ -a, \qquad a \leq 0 \end{cases},\quad \left| b \right| = \begin{cases} b, \qquad b \geq 0 \ -b, \qquad b \leq 0 \end{cases}$$ therefore for $a,b\ge 0: |a+b|= a+b = |a|+|b|$.     – Axion004 Oct 11 '21 at 23:13
  • @Owen Sorry for the delay... that's what surgery does. Thank you very much. Making a lot more sense. – user1115542 Oct 13 '21 at 18:42
  • @Axion004 That's very very nice. Thank you. – user1115542 Oct 13 '21 at 18:43

1 Answers1

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Try squaring both sides and noting $|a+b|^2 = (a+b)^2$

This is a classic approach

WhoDatBoy
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