1. (a) Prove that for all real numbers $a$ and $b$, $$|a| \le b \text{ iff } -b \le a \le b.$$

(b) Prove that for any real number $x$, $$-|x| \le x \le |x|.$$ (Hint: Use part (a).)

(c) Prove that for all real numbers $x$ and $y$, $$|x+y| \le |x| + |y|.$$ (This is called the triangle inequality. One way to prove this is to combine parts (a) and (b), but you can also do it by considering a number of cases.)

I am trying to prove part 12c using parts (a) and (b). I am already well aware that I can look up a proof for the triangle inequality by cases, but I would like to know how to solve it in the context of 12a and 12b. Any help would be appreciated!


Applying part $(b)$ we have that $$\begin{array}{l} -|x| \le x \le |x|\\ -|y| \le y \le |y|. \end{array}$$

Adding the $2$ inequalities we take: $$-(|x|+|y|) \le x+y \le |x|+|y|.$$ Try to apply part $(a)$ to the above inequality and see if you can get the triangle inequality.

By part $(a)$ we have $$\lvert x+y \rvert \le \big| \lvert x\rvert + \lvert y\rvert \big|.$$ However, $ \big| \lvert x\rvert + \lvert y\rvert \big|= \lvert x \rvert + \lvert y\rvert ,$ due to $\lvert x \rvert + \lvert y\rvert \ge 0.$

  • $\begingroup$ Thank you! Your answer to my question was clear and terse. I appreciated the spoiler tag on the application of part (a) in particular. $\endgroup$ Aug 26 '15 at 16:46
  • $\begingroup$ You're welcome! At least, I hope you reached the result on your own, without the help of the spoiler. $\endgroup$ Aug 26 '15 at 16:55
  • $\begingroup$ As an aside, the spoiler could use a small correction. The absolute value function should not be applied to the right side of the inequality since part (a) states that $|a| \le b$, not $|a| \le |b|$. On the bright side, this makes the solution even simpler! $\endgroup$ Aug 26 '15 at 20:33
  • $\begingroup$ Yes, indeed. As long as $b \ge 0.$ $\endgroup$ Aug 27 '15 at 5:27

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