# Absolute Value Theorem

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:

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

• It seems to me that you need $a + b \leq \big|a + b\big|$, if $x \leq y$ and $z \leq x$, then $z \leq y$. – Jared Jan 18 '15 at 0:16
• how do you write $-5 \le x \le 5$ using absolute vale symbol? – abel Jan 18 '15 at 0:18
• 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 $\big|a + b\big|$. – Jared Jan 18 '15 at 0:31
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 $-1$).
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.