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Thread: inequality

  1. #1
    key
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    inequality

    prove that |a+b|< |a|+|b| is true for all reall numbers
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  2. #2
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    Quote Originally Posted by key View Post
    prove that |a+b|< |a|+|b| is true for all reall numbers
    If $\displaystyle a\geq 0$ we end up with:
    $\displaystyle |a+b|\leq a + |b|$.
    Now if $\displaystyle -a\leq b <0$ we have:
    $\displaystyle a+b\leq a + |b| \implies b\leq |b|$ which is true.
    And if $\displaystyle 0\leq b$ we have:
    $\displaystyle a+b\leq a+b$ which is true.
    Now if $\displaystyle a<0$ let $\displaystyle a=-c$ where $\displaystyle c>0$ so:
    $\displaystyle |a+b|\leq |a|+|b| \implies |c+(-b)| \leq |c|+|-b|$.
    Which we established above already.
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