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Triangle inequality for a normalized absolute distance

  1. Sep 11, 2012 #1
    Hi, can you please give me some hints to show that
    [tex]\frac{|a-b|}{1+|a|+|b|} \leq \frac{|a-c|}{1+|a|+|c|}+\frac{|c-b|}{1+|c|+|b|}, \forall a, b, c \in \mathbb{R}.[/tex]
    I tried to get this from
    [tex]|a-b| \leq |a-c|+|c-b|, \forall a, b, c \in \mathbb{R},[/tex]
    but I couldn't succeed.

    Thank you.
     
  2. jcsd
  3. Sep 11, 2012 #2
    What do you get when you multiply by all the denominators?
    Can you do that and isolate |a-b| on the lhs and |a-c|+|c-b| on the rhs?
     
  4. Sep 11, 2012 #3
    I tried to play around with that approach but I couldn't get anything.
     
  5. Sep 11, 2012 #4

    jbunniii

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    I played around with this for a few minutes but didn't find a proof. However, there's a similar inequality that I do know how to prove. Perhaps you can adapt this proof, or use this inequality to prove yours.

    Claim:

    [tex]\frac{x}{1 + x} \leq \frac{y}{1 + y} + \frac{z}{1 + z}[/tex]

    for all non-negative x, y, z such that x <= y + z.

    Sketch of proof:

    First show that the function f(t) = t / (1 + t) is monotonically increasing for non-negative t. Then apply this fact to t1 = x and t2 = y + z.
     
  6. Sep 14, 2012 #5
    I still think you can definately do it by multiplying with the denominators, you just need to be very persistent and the calculation is very lengthy.
     
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