Bulgarian National Mathematical Olympiad

It follows from (i) that $f$ is strictly increasing function. Also, (ii) implies $$x+y\ge f(f(x+y)).(1)$$

Furthermore, (i) gives $f(f(x+y))\ge f(f(x)+y)$ and the substitution $x\to y$ and $y\to f(x)$ in (i) implies $f(f(x)+y)\ge f(x)+f(y)$. Since $f$ is increasing we have ${\lim }_{x\to 0^{+}}f(x)={\inf }_{x>0}f(x)=\ell \ge 0$. Note that (ii) implies ${\lim }_{x\to 0^{+}}f(f(x))=0$. Assume $\ell >0$. Since $f$ is increasing we have $f(f(x))\ge f(\ell )>0$, a contradiction to ${\lim }_{x\to 0^{+}}f(f(x))=0$. Therefore $\ell =0$ and ${\lim }_{x\to 0^{+}}f(x)=0$. Letting $y\to 0^{+}$ in (1) we obtain $x\ge f(x)$ for all positive $x$. It follows now from (i) that $$\begin{gathered} x+y\ge f(x+y)\ge f(x)+y, \\ x-f(x)\ge f(x+y)-f(x)-y\ge 0. \end{gathered}$$ Fix $x+y$ and let $x\to 0^{+}$ in the above inequalities. We have $f(x+y)=x+y$ meaning that $f(x)=x$ for all positive $x$. This function is obviously a solution to the problem.