I have finished the entire problem except the 5th part. Roughly speaking the questions asks to show that over $\mathbb{R}$ prove/disprove that the distance function

$$ d_5(x,y) = \frac{|x-y|}{1 + |x-y|} $$

Is a metric. I have proven symmetry and I have proven that if $d_5(x,y) = 0$ that $x = y$ so all that remains to be shown is the triangle inequality namely :

$$ \frac{|x-y|}{1 + |x-y|}+ \frac{|y-z|}{1 + |y-z|} \ge \frac{|x-z|}{1 + |x-z|} $$

This of course can be cast more simply in terms of $u,v$ in $\mathbb{R}$ as

$$ \frac{|u|}{1 + |u|}+ \frac{|v|}{1 + |v|} \ge \frac{|u+v|}{1 + |u+v|} $$

This reminds me an awful lot of contest style inequalities but i can't for the love of god place which one this is (not that it should matter, I should ideally be able to just generate a proof with ease).

My one angle of attack was to do case work on fixing $u,v$ to be both positive or both negative, or a mix of positive and negative. But is there a easier way to proceed than that?


1 Answer 1


\begin{align*} \dfrac{|u+v|}{1+|u+v|}&=1-\dfrac{1}{1+|u+v|}\\ &\leq1-\dfrac{1}{1+|u|+|v|}\\ &=\dfrac{|u|+|v|}{1+|u|+|v|}\\ &=\dfrac{|u|}{1+|u|+|v|}+\dfrac{|v|}{1+|u|+|v|}\\ &\leq\dfrac{|u|}{1+|u|}+\dfrac{|v|}{1+|v|}. \end{align*}

  • $\begingroup$ Very nice, so easy to follow, not so easy to come up with . $\endgroup$ Dec 14, 2017 at 7:11

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