Do we have the following inequality:

$$a-b\leq |a|+|b|$$

I have considered $4$ cases:

  • $a\leq0,b\leq0$

  • $a\leq0,b>0$

  • $a>0,b\leq0$

  • $a>0,b>0$

and see this inequality is true. However I want to make sure about that.


Use the triangle inequality: $$ a - b \leq \vert a - b \vert \leq \vert a \vert + \vert b \vert. $$

  • 3
    $\begingroup$ To expand on the second inequality: $|a-b|$ is the distance from $a$ to $b$ on the number line, and $|a|+|b|$ is the distance from $a$ to $0$ to $b$. $\endgroup$ Apr 14 '16 at 15:24

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

Above, we used the inequality $x \leq |x|$ twice: first with $x=a$ , and then with $x=-b$.


Right. Observe that $a\leq |a|$ and that $(-b)\leq |(-b)|=|b|.$ Adding, we have $a-b=a+(-b)\leq |a|+|(-b)|=|a|+|b|.$


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