proof of the inequality? prove that $$\frac{|x+y|}{1+|x+y|}\le\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|} $$
To solve this I assumed if $x,y$ are of opposite signs and  $x\gt y$ then  $|x+y|\le |x|$.
we get 
$$\frac{|x|}{1+|x|}\ge\frac{|x+y|}{1+|x+y|}$$
how to proceed further..
as answered by hans 
 to prove that the function $$f(x)=\frac{x}{1+x}$$ is monotonous ,$y\ge x$ then 
$$\frac{1}{1+y}\le\frac{1}{1+x}$$
$$1-\frac{1}{1+y}\ge 1-\frac{1}{1+x}$$
$$\frac{y}{1+y}\ge\frac{x}{1+x}$$
further $|x+y|\le|x|+|y|$ and using the above inequality 
One gets 
$$\frac{|x|+|y|}{1+|x|+|y|}\ge\frac{|x+y|}{1+|x+y|}$$
 A: This can be done in three steps:


*

*Prove or quote that $|x+y| \le |x| + |y|$.

*Prove that $f(z) = \frac{z}{z+1}$ is increasing on $[0,\infty)$. This can be done with Calculus or with an algebra step.

*Prove that $\frac{a+b}{1+a+b} \le \frac{a}{1+a} + \frac{b}{1+b}$ for all $a, \, b \ge 0$. That is also an algebra step.


Then put these three arguments together to make a proof. 
A: All you need is the triangle inequality: $\vert x+y\vert \leq \vert x\vert +\vert y\vert$. 
Now, if LHS=$0$, there is nothing to prove. Otherwise, we have
$$\frac{1}{\vert x\vert +\vert y\vert}\le\frac{1}{\vert x+y\vert }\Rightarrow 1+\frac{1}{\vert x\vert +\vert y\vert}\le 1+\frac{1}{\vert x+y\vert }\Rightarrow \frac{1}{1+\frac{1}{\vert x+y\vert }}\le \frac{1}{1+\frac{1}{\vert x\vert +\vert y\vert}}\Rightarrow \frac{\vert x+y\vert }{1+\vert x+y\vert }\le \frac{\vert x\vert +\vert y\vert }{1+\vert x\vert +\vert y\vert }=\frac{\vert x\vert}{1+\vert x\vert +\vert y\vert }+\frac{\vert y\vert }{1+\vert x\vert +\vert y\vert }\le \frac{\vert x\vert}{1+\vert x\vert }+\frac{\vert y\vert }{1+\vert y\vert }$$.
