Triangle inequality for a normalized absolute distance

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Homework Help Overview

The discussion revolves around proving a specific inequality involving normalized absolute distances among real numbers, specifically the expression \(\frac{|a-b|}{1+|a|+|b|} \leq \frac{|a-c|}{1+|a|+|c|}+\frac{|c-b|}{1+|c|+|b|}\) for all \(a, b, c \in \mathbb{R}\). Participants are exploring methods to establish this inequality based on known properties of absolute values and inequalities.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the possibility of manipulating the inequality by multiplying through by the denominators to isolate terms. There is also mention of a related inequality that might provide insight into the proof. Some participants express difficulty in applying these methods effectively.

Discussion Status

The discussion is ongoing, with participants sharing different approaches and insights. While some suggest that multiplying by the denominators could lead to a solution, others are exploring alternative inequalities that may be adapted for this proof. There is no clear consensus on the best method yet.

Contextual Notes

Participants note the complexity of the calculations involved and the need for persistence in working through the algebraic manipulations. There is also an acknowledgment of a similar inequality that may serve as a useful reference point.

buraq01
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Hi, can you please give me some hints to show that
\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}.
I tried to get this from
|a-b| \leq |a-c|+|c-b|, \forall a, b, c \in \mathbb{R},
but I couldn't succeed.

Thank you.
 
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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?
 
I tried to play around with that approach but I couldn't get anything.
 
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:

\frac{x}{1 + x} \leq \frac{y}{1 + y} + \frac{z}{1 + z}

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.
 
I still think you can definitely do it by multiplying with the denominators, you just need to be very persistent and the calculation is very lengthy.
 

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