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?

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 $\int\frac{\vert f-h\vert}{1+\vert f-h\vert}\leq\int\frac{\vert f-g\vert}{1+\vert f-g\vert}+\int\frac{\vert g-h\vert}{1+\vert g-h\vert}$? I tried C-S: $$\sum_{cyc}\frac{|x|}{1+|x|}=\sum_{cyc}\frac{x^2}{|x|+x^2}\geq\frac{(x+y)^2}{x^2+y^2+|x|+|y|}.$$ Thus, it's enough to prove that $$\frac{(x+y)^2}{x^2+y^2+|x|+|y|}\geq\frac{(x+y)^2}{|x+y|+(x+y)^2}$$ or $$|x+y|+(x+y)^2\geq x^2+y^2+|x|+|y|,$$ which is wrong for $(x,y)=(1,-1).$

Solution 1:

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

Solution 2:

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.

Solution 3:

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:

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

But:

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