$$x,y \epsilon R,\frac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$$ I have to prove this inequality.I used the triangle inequality once and have this step. $$\frac{|x+y|}{1+|x+y|}\le\frac{|x|+|y|}{1+|x+y|}$$ Now how can I prove the rest? I don't have any idea how to use triangle inequality for the rest of the proof.
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http://math.stackexchange.com/questions/788995/proving-a-metric-with-absolute-value, http://math.stackexchange.com/questions/355493/defining-a-metric-space – Martin R Apr 22 '16 at 12:49
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Note $f(t): t \mapsto \frac{t}{1+t}$ is increasing for $t> 0$, and so given $|x+y| \leqslant |x|+|y|$,
$$\frac{|x+y|}{1+|x+y|} \leqslant \frac{|x|+|y|}{1+|x|+|y|}= \frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|} \leqslant \frac{|x|}{1+|x|}+\frac{|y|}{1+|y|} $$
Equality is possible when $xy=0$.
Macavity
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Oops! Sorry, I just looked at the chain of inequalities. I'll delete my comment. – Bernard Apr 22 '16 at 12:06
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Obviously, $\frac{|x|+ |y|}{1+ |x+ y|}= \frac{|x|}{1+ |x+ y|}+ \frac{|y|}{1+ |x+ y|}$. Now, what can you say about the relative sizes of 1+ |x+ y| and 1+ |x| and 1+ |y|?
user247327
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but we don't know that both x and y have same sign or different signs. so i think we cant say that $1+|x+y|\gt 1+|x| or 1+|y|$ – Anju Cheran Apr 22 '16 at 11:54