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Triangle Inequality for integrals proof

  1. Oct 16, 2011 #1
    1. The problem statement, all variables and given/known data
    What I want to show is this:
    ∫|x+y| ≤ ∫|x| + ∫|y|

    2. Relevant equations
    |x+y| ≤ |x| + |y|


    3. The attempt at a solution

    So I thought if I used the triangle inequality I could get to something along the lines of:

    Lets g belong to the real numbers
    ∫|x+y| = ∫|x+g-g+y|≤ ∫|x+g| + |y-g|= ∫|x+g| + ∫|y-g|

    As g belongs to the reals it can be zero meaning ∫|x+y| ≤ ∫|x| + ∫|y|.

    Now the problem with this is that is uses the triangle inequality and I have no idea if the triangle inequality works this way, and if it does I need to prove it, and I have no idea about where to start that from.
     
  2. jcsd
  3. May 13, 2013 #2

    Ackbach

    User Avatar
    Gold Member

    I suppose I am necro-posting here, but this result follows from two facts. One is the linearity of integration:
    [tex]\int (f(x)+g(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx,[/tex]
    and the other is that integration preserves inequalities: if [itex]f(x) \le g(x)[/itex] on the interval [itex] [a,b] [/itex], then
    [tex] \int_{a}^{b}f(x) \, dx \le \int_{a}^{b} g(x) \, dx. [/tex]
     
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