# Prove an inequality $\frac{|x+y|}{1+|x+y|} \le \frac{|x|}{1+|x|} + \frac{|y|}{1+|y|}$ for $x,y \in \mathbb{R}$ [duplicate]

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

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

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

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