<s>[INST] 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?
 [/INST] 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 ...
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