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Triangle inequality for a normalized absolute distance

  • Thread starter buraq01
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  • #1
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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.
 

Answers and Replies

  • #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?
 
  • #3
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I tried to play around with that approach but I couldn't get anything.
 
  • #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.
 
  • #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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