<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] Let $f(x)=\frac{x}{1+x}.$
Then by Jensen inequality
$$
f(a)+f(b)=\frac{a}{1+a}+\frac{b}{1+b} \geq 2 f( \frac{a+b}{2})=\frac{a+b}{1+\frac{a+b}{2}} \geq \frac{a+b}{1+a+b}.
$$
</s>