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?
 A: We get $$\frac{a}{1+a}+\frac{b}{1+b}-\frac{a+b}{1+a+b}={\frac {ba \left( a+b+2 \right) }{ \left( 1+a \right)  \left( 1+b
 \right)  \left( 1+a+b \right) }}
\geq 0$$
the numerator can be calculated as
$$a(1+b)(1+a+b)+b(1+a)(1+a+b)-(a+b)(1+a)(1+b)=...$$
we also have under the same conditions
$$\frac{a+b+c}{1+a+b+c}\le \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}$$ and so on ...
A: Let $f(x)=\frac{x}{1+x}.$
Then by Jensen inequality
$$
f(a)+f(b)=\frac{a}{1+a}+\frac{b}{1+b} \geq 2 f( \frac{a+b}{2})=\frac{a+b}{1+\frac{a+b}{2}} \geq \frac{a+b}{1+a+b}.
$$
A: $$
\frac{a+b}{1+a+b} =\frac{a}{1+a+b} + \frac{b}{1+a+b}
$$
then prove
$$
\frac{a}{1+a+b} \leq \frac{a}{1+a} 
$$
A: 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.
A: I guess you're almost there...
If $a\ge 0$, then $1+a\ge 1$ and so $(1+a)(1+b)\ge 1+b$. This gives you
$$\frac b {(1+a)(1+b)}\le \frac b {1+b}$$
and the result you're looking for follows.
A: Since both $a$ and $b$ are nonnegative so are ${{1}\over{1+a}}$ and ${{b}\over{1+b}}+a$ and their multiplication.
