Prove that : $\frac{|a+b|}{1+|a+b|} \leq \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$ Exercise :

Prove that :
  $$\frac{|a+b|}{1+|a+b|} \leq \frac{|a|}{1+|a|} + \frac{|b|}{1+|b|}$$
  for $a,b \in \mathbb R$.

Methods I have tried so far include:
Using the triangle inequality on the numerator on the left side, but I got an expression which was sometimes too big, so it's impossible.
Using a similar method on the right side, but I got a pretty nasty expression so I don't think that's the way.
Going case by case for every pair of a,b but this is also very long.
Can someone give a hint?
 A: The pedestrian approach works: with $x=|a|$, $y=|b|$, and $z=|a+b|$, we have
$$
(x(1+y)+y(1+x))(1+z)-z(1+x)(1+y)=[x+y-z]+2[xy+xyz]\geq0.
$$
The last inequality above holds because $x+y-z\geq 0$ (triangle inequality) and because $x,y,z\geq 0$.
A: Let $s=|a|$, and let $t=|b|$.

Consider two cases . . .

Case $(1)$:$\;ab\ge 0$.

Since $ab\ge 0$, it follows that $a,b$ do not have opposite signs, hence  $|a+b|=s+t$, so

\begin{align*}
&\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}-\frac{|a+b|}{1+|a+b|}\\[4pt]
=\;&\frac{s}{1+s}+\frac{t}{1+t}-\frac{s+t}{1+(s+t)}\\[4pt]
=\;&\frac{st(2+s+t)}{(1+s)(1+t)(1+s+t)}\\[4pt]
\ge\;&\;0\\[4pt]
\end{align*}
Thus, the claimed inequality holds for case $(1)$.

Case $(2)$:$\;ab < 0$.

Without loss of generality, assume $s\ge t$.

Since $ab < 0$, it follows that $a,b$ have opposite signs, hence $|a+b|=s-t$, so
\begin{align*}
&\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}-\frac{|a+b|}{1+|a+b|}\\[4pt]
=\;&\frac{s}{1+s}+\frac{t}{1+t}-\frac{s-t}{1+(s-t)}\\[4pt]
=\;&\frac{t\bigl(s(s-t)+2s+2\bigr)}{(1+s)(1+t)(1+(s-t))}\\[4pt]
\ge\;&\;0\\[4pt]
\end{align*}
Thus, the claimed inequality also holds for case $(2)$.

Therefore the claimed inequality always holds.
A: The required inequality is equivalent to
$$1+\frac{1}{1+|a+b|}\geq \frac{1}{1+|a|}+\frac{1}{1+|b|}\,.$$
By the Triangle Inequality, $|a+b|\leq |a|+|b|$, so we have
$$1+\frac{1}{1+|a+b|}\geq 1+\frac{1}{1+|a|+|b|}\,.$$
Therefore, it suffices to show that
$$1+\frac{1}{1+|a|+|b|}\geq \frac{1}{1+|a|}+\frac{1}{1+|b|}\,.$$
This is clearly true, as
$$\begin{align}1+\frac{1}{1+|a|+|b|}&=\frac{2+|a|+|b|}{1+|a|+|b|}\geq \frac{2+|a|+|b|}{1+|a|+|b|+|ab|}\\&=\frac{2+|a|+|b|}{\big(1+|a|\big)\big(1+|b|\big)}=\frac{1}{1+|a|}+\frac{1}{1+|b|}\,.\end{align}$$
Hence,
$$\frac{|a+b|}{1+|a+b|}\leq \frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}\,,$$
as desired.  The equality holds if and only if $a=0$ or $b=0$.
