Is $\frac{\vert x+y\vert}{1+\vert x+y\vert}\leq\frac{\vert x\vert}{1+\vert x\vert}+\frac{\vert y\vert}{1+\vert y\vert}$ true for all $x,y\in\mathbb{R}$?

If not, how can I prove that $\int\frac{\vert f-h\vert}{1+\vert f-h\vert}\leq\int\frac{\vert f-g\vert}{1+\vert f-g\vert}+\int\frac{\vert g-h\vert}{1+\vert g-h\vert}$?

  • $\begingroup$ Hint: $\frac{x}{x+1} = 1-\frac{1}{x+1}$ $\endgroup$ – flawr Oct 20 '14 at 19:21

Hint Consider the function $f(x)=\frac{x}{1+x}$. Show that the function is increasing everywhere.

Now comes the interesting part.

Clearly, by Triangle inequality $|x+y|\le |x|+|y|$

Then $f(|x+y|)\le f(|x|+|y|)$

Now play with the R.H.S. and get the inequality.


Consider function $f(z)=\frac{z}{1+z}$. You can check that $f'(z)=\frac{1}{(z+1)^2}$, so it's increasing function. So for $a \leq b$ you have:

$$f(a) \leq f(b)$$

By triangle inequality $|x+y| \leq |x|+|y|$, so:

$$f(|x+y|) \leq f(|x|+|y|)$$


$$f(|x|+|y|)=\frac{|x|+|y|}{1+|x|+|y|}=\frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|} \leq \frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}$$


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