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I'm working on a proof and I need to show that $|a|>|b|/2$ knowing that $|a-b|<|b|/2$. I would like to do it without enumerating the different cases.

Thanks for you ideas.

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Using the triangle inequality in the 2nd inequality below: $$ \frac{|b|}{2}>|a-b|=|b-a|\geq|b|-|a|\implies |a|>|b|-\frac{|b|}{2}=\frac{|b|}{2}. $$

Edit: a few more details: the triangle inequality states that: $$ |x+y|\leq|x|+|y|. $$ So applying it with $x=b-a$ and $y=a$ yields $$ |b|=|b-a+a|\leq |b-a|+|a|\implies|b-a|\geq|b|-|a|. $$

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  • $\begingroup$ In fact you use the reverse triangle inequality $|a-b|\geq |(|a|-|b|)|$ but how did you drop the outside absolute value of the right hand side? $\endgroup$ Nov 26, 2014 at 22:30
  • $\begingroup$ OK, thanks for your edit. $\endgroup$ Nov 26, 2014 at 22:41

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