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This is a problem from Spivak's Calculus $3^{rd}$ ed., Chapter I, Problem $12$(vii)

Indicate when $|x + y + z| = |x| + |y| + |z|$ holds, and prove your statement.

My attempt: Clearly, $|x + y + z| \leq |x + y| + |z| \leq |x| + |y| + |z|$, so since $|x+y+z| = |x| + |y| + |z|$, it follows that $|x + y + z|$ = $|x + y| + |z|$ and $|x + y| = |x| + |y|$. Since $|x + y| = |x| + |y|$, we can conclude that $x$ and $y$ have the same sign or one of the two is 0. Since $|x + y + z|$ = $|x + y| + |z|$, $x+y$ and $z$ have the same sign or one of the two is 0. Therefore, one of the following holds:

  1. two of the $x,y,z$ are 0
  2. one of the $x,y,z$ is 0 and the other two have the same sign
  3. $x,y,z$ have the same sign.

Is my proof all right? How should I attack this problem? Any general tips?

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1 Answer 1

up vote 3 down vote accepted

You are correct, but there is a simpler approach. This could be generalized to $$\left| \sum_k a_k \right| = \sum_k |a_k|$$ if and only if all non-zero elements have the same sign. To prove it, start with the fact, that when $a_i = 0$, then it does not change anything. Follow with small lemma

$$x < 0 < y \quad \implies \quad |x+y| \color{red}{<} |x| + |y|\tag{$\color{red}{\spadesuit}$}$$

and prove, that if $a_i < 0 < a_j$ for some $i$ and $j$, then $$\left|\sum_{k} a_k\right| \leq | a_i + a_j| + \left|\sum_{k, k\neq i, k\neq j} a_k\right| \ \color{red}{\stackrel{\spadesuit}{\mathbf{<}}}\ |a_i| + |a_j| + \left|\sum_{k, k\neq i, k\neq j} a_k\right| \leq \sum_{k} |a_k|.$$

I hope this helps $\ddot\smile$

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Nice approach! Thank you :-) –  Stavros Aug 21 '13 at 11:15

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