How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab}
$$
which gives me
$$
\frac{a}{1+a} + \frac{b}{(1+a)(1+b)}
$$
How can I reduce the second term further, and get the required result?

In order to prove that $f(x)=\frac{x}{1+x}$ is sublinear on $\mathbb{R}^+$ it is enough to notice that $f'(x)=\frac{1}{(1+x)^2}$ leads to
$$ \frac{f(a+b)-f(a)}{f(b)-f(0)} = \frac{\int_{a}^{a+b}\frac{dx}{(1+x)^2}}{\int_{0}^{b}\frac{dx}{(1+x)^2}}\leq 1 $$
since $f'(x)$ is decreasing. In other terms, the sublinearity is a consequence of the concavity.
