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This is from a book. I don't understand how the inequalities are combined to one inequality. Are they added/subtracted?

$z_1$ and $z_2$ are complex numbers. We have the inequalities $$\lvert z_2\rvert -\lvert z_1\rvert \leq \lvert z_2-z_1\rvert $$ $$\lvert z_1\rvert -\lvert z_2\rvert \leq \lvert z_1-z_2\rvert $$ Combining these inequalities gives $$\lvert \lvert z_1\rvert -\lvert z_2\rvert \rvert \leq \lvert z_1-z_2\rvert$$

Attempt

If I add them: \begin{align} \lvert z_2\rvert -\lvert z_1\rvert + \lvert z_1\rvert -\lvert z_2\rvert &\leq \lvert z_2-z_1\rvert +\lvert z_1-z_2\rvert \\ 0&\leq \lvert z_2-z_1\rvert +\lvert z_1-z_2\rvert \\ 0&\leq 2\lvert z_1-z_2\rvert \end{align} Or if I subtract them: \begin{align} \lvert z_2\rvert -\lvert z_1\rvert -(\lvert z_1\rvert -\lvert z_2\rvert)&\leq \lvert z_2-z_1\rvert -( \lvert z_1-z_2\rvert )\\ 2\lvert z_2\rvert-2\lvert z_1\rvert &\leq \lvert z_2-z_1\rvert -\lvert z_1-z_2\rvert\\ 2\lvert z_2\rvert-2\lvert z_1\rvert &\leq 0 \end{align}

I'm stuck here!

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For $a,b\in\Bbb R$, it holds $\lvert a-b\rvert=\begin{cases} b-a&\text{if }b\ge a\\ a-b&\text{if }a>b\end{cases}$.

In the first case you use one inequality, in the other you use the other one (obviously, $\lvert z_1-z_2\rvert=\lvert z_2-z_1\rvert$).

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  • $\begingroup$ Subject says complex numbers. $\endgroup$ – MvG Apr 2 '17 at 15:11
  • $\begingroup$ @mvg Yeah, but he needs only the real version. $\endgroup$ – user228113 Apr 2 '17 at 15:13
  • $\begingroup$ I should have thought about this for one more second before commenting. Nevertheless, it would be interesting to not rely on this well-known fact, but only on the given properties if possible. $\endgroup$ – MvG Apr 2 '17 at 15:14
  • $\begingroup$ @MvG How is the definition of absolute value of a real number not a given property when this appears in the problem statement? $\endgroup$ – mathematician Apr 2 '17 at 15:20
  • $\begingroup$ @mathematician The $\lvert z\rvert$ in the problem is the complex modulus $\sqrt{z\overline z}$, which can, in principle, be applied to real numbers as well (aka $\sqrt{x^2}$). $\endgroup$ – user228113 Apr 2 '17 at 15:21
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We have $|x|\leq y$ iff $-y\leq x\leq y$ and with $$\lvert z_2\rvert -\lvert z_1\rvert \leq \lvert z_2-z_1\rvert $$ $$\lvert z_1\rvert -\lvert z_2\rvert \leq \lvert z_1-z_2\rvert $$ have $$-\lvert z_1-z_2\rvert\leq \lvert z_1\rvert -\lvert z_2\rvert \leq \lvert z_1-z_2\rvert $$ iff $$\Big|\lvert z_1\rvert -\lvert z_2\rvert\Big| \leq \lvert z_1-z_2\rvert $$

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Hint:

$$|z_1-z_2| = |z_2-z_1|$$

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  • 1
    $\begingroup$ Wouldn't this be suitable for a Comment? The OP did a good bit of thinking, and such a brief remark in lieu of a full explanation seems to fall short of the requirements of an Answer. $\endgroup$ – hardmath Apr 2 '17 at 16:20

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