Belarusian Mathematical Olympiad

Answer: such function doesn't exist. Suppose that such a function exists. Putting $x=1$ in the inequality $$f(x+y)\ge yf(x)+f(f(x))(1)$$ we obtain $f(1+y)\ge ay+b$ where $a=f(1)>0$, $b=f(f(1))>0$. So $$f(z)\ge az+b-a(2)$$ for all $z>1$. From (2) it follows that $f(x)>1$ for all sufficiently large $x$ (namely when $x\ge \frac{1-b+a}{a}$). Putting $x=y=\frac{z}{2}$ in (1) we get $$f(z)\ge \frac{z}{2}\cdot f\left(\frac{z}{2}\right)+f\left(f\left(\frac{z}{2}\right)\right)>\frac{z}{2}\cdot f\left(\frac{z}{2}\right)\ge \frac{z}{2}\left(a\cdot \frac{z}{2}+b-a\right)=\frac{a}{4}\cdot z^2+\frac{b-a}{2}\cdot z.$$ Therefore $$f(z)>\frac{a}{5}\cdot z^2>z(3)$$ for all $z$ large enough. Further, we have $$f(x+y)\ge yf(x)+f(f(x))>f(f(x))>f(x)$$ for all $x$ large enough. Hence $f(x)$ is an increasing function for $x$ large enough. Therefore, from $f(x+y)>f(f(x))$ it follows that $x+y>f(x)$ for all $y>0$ and all $x$ large enough. Together with (3) this leads $x+y>\frac{a}{5}\cdot x^2$ for all $x$ large enough, which is obviously wrong. The contradiction shows that there is no such function.