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My only version is to use properties of function $f(x)=\frac{|x|}{1+|x|}$.

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marked as duplicate by Carl Mummert, kimchi lover, Martin R, Community Jan 7 '18 at 19:31

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$$\frac{|x+y|}{1+|x+y|}=1-\frac{1}{1+|x+y|}\leq1-\frac{1}{1+|x|+|y|}=$$ $$=\frac{|x|+|y|}{1+|x|+|y|}=\frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|}\leq$$ $$\leq\frac{|x|}{1+|x|}+\frac{|y|}{1+|y||}.$$

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  • $\begingroup$ and now the general case! $\endgroup$ – Dr. Sonnhard Graubner Jan 7 '18 at 17:38
  • $\begingroup$ In the general case it's the same. $\endgroup$ – Michael Rozenberg Jan 7 '18 at 17:39

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