# inequality

• Sep 9th 2007, 05:33 AM
key
inequality
prove that |a+b|< |a|+|b| is true for all reall numbers
• Sep 9th 2007, 06:22 AM
ThePerfectHacker
Quote:

Originally Posted by key
prove that |a+b|< |a|+|b| is true for all reall numbers

If $a\geq 0$ we end up with:
$|a+b|\leq a + |b|$.
Now if $-a\leq b <0$ we have:
$a+b\leq a + |b| \implies b\leq |b|$ which is true.
And if $0\leq b$ we have:
$a+b\leq a+b$ which is true.
Now if $a<0$ let $a=-c$ where $c>0$ so:
$|a+b|\leq |a|+|b| \implies |c+(-b)| \leq |c|+|-b|$.