Triangle Inequality question with fractions Given $a, b \in \mathbb{R}$, prove that 
$$\frac{|a + b|}{1 + |a + b|} \le \frac{|a|}{1 + |a|} + \frac{|b|}{1+|b|}$$
When does equality hold?
The only useful thing I could get (using the triangle inequality) is:
$$\begin{align}1 + \left|\left(a + \frac{ab}{2}\right) + \left(b - \frac{ab}{2}\right)\right| &\le 1 + \left|a + \frac{ab}{2}\right| + \left|b - \frac{ab}{2}\right|\\
&\le 1 + |a| + |b| + |ab|\\
&=(1 + |a|)(1 + |b|)\end{align}$$
But it seems to be inapplicable! Any ideas?

Because the function $f(x)=\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x}$ is increasing, we have
$$\frac{|a+b|}{1+|a+b|} = \frac{1}{1+\frac{1}{|a+b|}} \le \frac{1}{1+\frac{1}{|a|+|b|}}=\frac{|a|}{1+|a|+|b|}+\frac{|b|}{1+|a|+|b|}\le\frac{|a|}{1+|a|}+\frac{|b|}{1+|b|}$$
In the first inequality we have used the triangle inequality $|a+b|\le |a|+|b|$ and in the second inequality, the positivity of $|a|$ and $|b|$.
That is, equality holds if and only if $a=0$ or $b=0$ because otherwise the last inequality is strict.
