## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 1 - Precalculus Review - 1.1 Real Numbers, Functions, and Graphs - Exercises - Page 12: 82

#### Answer

See the proof below.

#### Work Step by Step

Since $-|a| \le a \le |a|$ and $-|b| \le b \le |b|$, then by addition we have $-|a|+-|b|\le a+b\le |a|+|b|$ $-(|a|+|b|)\le a+b\le |a|+|b|$ This can be rewritten as: $|a+b|\le |a|+|b|$ which is the triangle inequality.

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