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Triangle inequality for a normalized absolute distance
I tried to play around with that approach but I couldn't get anything.- buraq01
- Post #3
- Forum: Calculus and Beyond Homework Help
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B
Triangle inequality for a normalized absolute distance
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.- buraq01
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- Absolute Inequality Triangle Triangle inequality
- Replies: 4
- Forum: Calculus and Beyond Homework Help