Case I: $a$ and $b$ have the same sign:
Let the function $f$ be defined by: $f(x) = \frac{\left | x \right |}{1+\left | x \right |}$. Obviously $f$ is even, and $f’(x)$ is not defined in $x=0$, but $f$ is differentiable in the two domains $\mathbb{R}_-$ and $\mathbb{R}_+$, and we have by inspection: $f’’(x) < 0$ in both domains. Thus $f$ is concave on both sides of the ordinate.
Jensens inequality with equal weights then gives us:
\[f\left ( \frac{a+b}{2} \right ) \leq \frac{1}{2}\left ( f(a) + f(b)\right ) \\ \frac{\frac{1}{2}\left | a+b \right |}{1+\frac{1}{2}\left | a+b \right |} \leq \frac{1}{2}\left ( \frac{\left | a \right |}{1+\left | a \right |} +\frac{\left | b \right |}{1+\left | b \right |} \right )\] - or
\[\frac{\left | a+b \right |}{1+\frac{1}{2}\left | a+b \right |} \leq \frac{\left | a \right |}{1+\left | a \right |} +\frac{\left | b \right |}{1+\left | b \right |}\], which immediately implies:
\[\frac{\left | a+b \right |}{1+\left | a+b \right |} \leq \frac{\left | a \right |}{1+\left | a \right |} +\frac{\left | b \right |}{1+\left | b \right |}\].
Case II: $a$ and $b$ have opposite sign. Here, we cannot use the concavity argument, but the inequality is still valid:
First note, that: $\left | a+b \right |\leq max\left \{ \left | a \right |,\left | b \right | \right \}$
WLOG let $\left | b \right | \geq \left | a+b \right |$. Denote $x= \left | a+b \right |, \Delta = \left | b \right |-x \geq 0$:
The inequality: $\frac{\left | a+b \right |}{1+\left | a+b \right |}=\frac{x}{1+x} \leq \frac{x+\Delta }{1+x+\Delta } = \frac{\left | b \right |}{1+\left | b \right |}$ is true because:
\[\frac{x+\Delta }{1+x+\Delta }-\frac{x}{1+x}= \frac{(1+x)(x+\Delta )-x(1+x+\Delta )}{(1+x)(1+x+\Delta )}=\frac{\Delta }{(1+x)(1+x+\Delta )}\geq 0\]. Thus the inequality holds, from which we immediately have:
\[\frac{\left | a+b \right |}{1+\left | a+b \right |}\leq \frac{\left | b \right |}{1+\left | b \right |}\leq \frac{\left | a \right |}{1+\left | a \right |}+\frac{\left | b \right |}{1+\left | b \right |}\].