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When trying to prove the inequality

$$ |a +b| \leq |a| + |b| \text{, for any real numbers a and b} $$ I manage to use the absolute value definition to get to following inequality:

$$ -\big(|a|+|b|\big) \leq a + b \leq |a| + |b| $$

However, the text book leaps foward and states that:

$$ \Big\{-\big(|a|+|b|\big) \leq a + b \leq |a| + |b|\Big\} \leftrightarrow \Big\{ |a + b| \leq |a| + |b|\Big\} $$

How did it jump to that conclusion?

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  • $\begingroup$ It seems to me that you need $a + b \leq \big|a + b\big|$, if $x \leq y$ and $z \leq x$, then $z \leq y$. $\endgroup$ – Jared Jan 18 '15 at 0:16
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    $\begingroup$ how do you write $-5 \le x \le 5$ using absolute vale symbol? $\endgroup$ – abel Jan 18 '15 at 0:18
  • $\begingroup$ Why not squaring both sides? $\endgroup$ – Vincenzo Oliva Jan 18 '15 at 0:25
  • $\begingroup$ Jared, I can get to a+b<=|a+b|, however with a+b<=|a|+|b| , how can I tell that |a+b| <= |a|+|b|? At most I can only prove the equality, not the inequality part. $\endgroup$ – MBdr Jan 18 '15 at 0:29
  • $\begingroup$ There are two cases: $a + b \geq 0$ in which case $a + b = \big|a + b\big|$ or $a + b < 0$ in which case $a + b = -\big|a + b\big|$ and thus $a + b < \big|a + b\big|$ (when the sum is negative). Therefore either $a + b$ equals $\big|a + b\big|$ or $a + b$ is less than $\big|a + b\big|$. $\endgroup$ – Jared Jan 18 '15 at 0:31
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The definition of absolute value is: $$|x| = \begin{cases} x & \text{if $x\geq 0$} \\ -x & \text{if $x<0$.} \end{cases}$$ So assume $a+b\geq 0$. Then $|a+b| = a+b\leq |a|+|b|$ by the inequality you've shown. If $a+b<0$, then $a+b = -|a+b|$, so $-(|a|+|b|)\leq -|a+b| \Longleftrightarrow |a+b|\leq |a|+|b|$ (the inequality flips since we divide by $-1$).

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so you have $-|a| - |b| \le a + b \le |a| + |b|$ that is $(a+b)$ in magnitude is less or equal to the nonnegative quantity $|a| +|b|$ writing this using absolute value notation is $$ |a +b | \le |a| +|b| $$ called the triangle inequality.

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