Proof for inequality with absolute values: $\frac{|x+y|}{1+|x+y|} \leq \frac{|x|}{1+|x|} + \frac{|y|}{1+|y|}$

If $x$ and $y$ are two real numbers then

$$\dfrac{|x+y|}{1+|x+y|} \leq \dfrac{|x|}{1+|x|} + \dfrac{|y|}{1+|y|}$$

How to prove above relation?

Hint:

Let $f(t)=\frac{t}{1+t}$ for $t \geq 0$. Then $$f^{'}(t)=\frac{1}{(1+t)^2} > 0 \qquad \forall t \geq 0.$$ Thus $f$ is an increasing function.

Now show that $$f(a+b) \leq f(a)+f(b) \qquad \forall a,b \geq 0.$$

$$\frac{|x|}{1+|x|}+\frac{|y|}{1+|y|}\geq\frac{|x|}{1+|x|+|y|}+\frac{|y|}{1+|x|+|y|}=$$ $$=\frac{|x|+|y|}{1+|x|+|y|}=1-\frac{1}{1+|x|+|y|}\geq1-\frac{1}{1+|x+y|}=\frac{|x+y|}{1+|x+y|}.$$

Let $f(t) = {\large{\frac{t}{1+t}}}$.

Then $f(t) = 1- {\large{\frac{1}{1+t}}}$, hence $f$ is increasing on $[0,\infty)$.

It follows that the LHS of the inequality doesn't get smaller if $x,y$ are replaced by $|x|,|y|$.

And clearly, if $x,y$ are replaced by $|x|,|y|$, the RHS stays the same.

Thus, it suffices to consider the case $x,y \ge 0$, since the truth of the inequality for that case implies the truth for the others.

But then, assuming $x,y \ge 0$, we have \begin{align*} &\,\frac{|x|}{1+|x|} + \frac{|y|}{1+|y|}-\frac{|x+y|}{1+|x+y|}\\[6pt] =&\;f(x)+f(y)-f(x+y)\\[6pt] =&\;\frac{(xy)(2+x+y)}{(1+x)(1+y)(1+x+y)} \qquad\text{(identically)}\\[6pt] \end{align*} which is obviously nonnegative.