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An inequality

  1. May 18, 2013 #1
    For arbitrary positive numbers ##\epsilon## and ##\delta## we know that ##0<\delta<\epsilon## such that ##0<A<B(\epsilon-\delta)+\epsilon C## for A, B, C>0. Can we conclude ##A<B(\epsilon-\delta)##?
     
    Last edited by a moderator: May 18, 2013
  2. jcsd
  3. May 18, 2013 #2

    Office_Shredder

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    When you say arbitrary, do you mean it holds for every choice of epsilon and delta, or there exists some epsilon and delta? Because for the latter it's clearly not true, and for the former no such numbers A,B,C exist. For example if [itex] \delta = \epsilon/2[/itex] you get
    [tex] 0<A< \epsilon( B/2+C) [/tex]
    And sending epsilon to 0 shows that A has to be zero which is a contradiction to it being positive
     
  4. May 18, 2013 #3
    Thanks for your comment. I mean that the above tow inequalities hold for for every choice of epsilon and delta.
     
  5. May 18, 2013 #4

    Mark44

    Staff: Mentor

    Here at Physics Forums, surround your expressions with pairs of $$ or ##. A single $ doesn't do anything.
     
  6. May 18, 2013 #5
    Thank you Mark44
     
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