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Inequality problem

  1. Aug 1, 2013 #1
    1. The problem statement, all variables and given/known data

    Given:

    [itex]|x-y| < K[/itex]

    [itex]x+y > K - 2[/itex]

    [itex]0 < K < 1[/itex]

    Prove:

    [itex]\frac{|1-K+x|}{|1+y|} < 1[/itex]







    3. The attempt at a solution

    I have tried using the fact that [itex]|x-y| < K \Rightarrow -K < x-y < K \Rightarrow y-K < x < y+K[/itex] to write [itex]\frac{1-K+x}{1+y} < \frac{1+y}{1+y} = 1[/itex]

    But I can't figure out how to show that the absolute value is less than one.

    I have also been trying various applications of the triangle inequality with little success.

    Any help would be greatly appreciated.
     
    Last edited: Aug 1, 2013
  2. jcsd
  3. Aug 1, 2013 #2

    jbunniii

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    Hint: what does the given inequality ##x + y > K - 2## tell you about ##1 - K + x##?
     
  4. Aug 2, 2013 #3
    It tells me that ##1 - K + x > -(y+1)##

    which I actually had written on my paper before, but thanks to you I think I've realized the connection...


    Since I already had from the first assumption ##1-K+x<1+y##, I now have

    ##-(y+1)<(1-K+x)<(y+1)##, which seems to imply ##|1-K+x| < (y+1)## as desired.

    Since we didn't know before that ##|y+1| = y+1##, it seems that this claim actually follows from the assumptions? I think it does since I think

    ##-A<T<A \Rightarrow A>0##, if both inequalities are satisfied, right?

    Thank you so much for your help.
     
  5. Aug 2, 2013 #4

    jbunniii

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    Yes, this implication is correct. Ignoring the ##T## in the middle of the inequality chain, you have ##-A < A##, so ##2A > 0##, hence ##A > 0##.

    Note that the fact that ##y + 1 > 0## is important, because you are dividing both sides of each inequality by this expression. If ##y+1## was zero, the division would be undefined, and if ##y+1## was negative, the direction of the inequality would flip. Fortunately both of these possibilities are excluded by ##y + 1 > 0##.
     
  6. Aug 2, 2013 #5
    That makes sense to me. Thanks again for the help.
     
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