<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] 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.
</s>