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?

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.