IMO Problem Shortlist

Assume that $$f(x-f(y))\le yf(x)+x\text{ for all real }x,y.\text{\hspace{1em}(1)}$$ Let $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a)\le x$ for all real $x$ and, equivalently, $$f(y)\le y+a\text{ for all real }y.\text{\hspace{1em}(2)}$$ Setting $x=f(y)$ in (1) yields in view of (2) $$a=f(0)\le yf(f(y))+f(y)\le yf(f(y))+y+a.$$ This implies $0\le y(f(f(y))+1)$ and thus $$f(f(y))\ge -1\text{ for all }y>0.\text{\hspace{1em}(3)}$$ From (2) and (3) we obtain $-1\le f(f(y))\le f(y)+a$ for all $y>0$, so $$f(y)\ge -a-1\text{ for all }y>0\text{\hspace{1em}(4)}$$ Now we show that $$f(x)\le 0\text{ for all real }x.\text{\hspace{1em}(5)}$$ Assume the contrary, i.e. there is some $x$ such that $f(x)>0$. Take any $y$ such that $$y<x-a\text{ and }y<\frac{-a-x-1}{f(x)}.$$ Then in view of (2) $$x-f(y)\ge x-(y+a)>0$$ and with (1) and (4) we obtain $$yf(x)+x\ge f(x-f(y))\ge -a-1,$$ whence $$y\ge \frac{-a-x-1}{f(x)}$$ contrary to our choice of $y$. Thereby, we have established (5). Setting $x=0$ in (5) leads to $a=f(0)\le 0$ and (2) then yields $$f(x)\le x\text{ for all real }x.\text{\hspace{1em}(6)}$$ Now choose $y$ such that $y>0$ and $y>-f(-1)-1$ and set $x=f(y)-1$. From (1), (5) and (6) we obtain $$f(-1)=f(x-f(y))\le yf(x)+x=yf(f(y)-1)+f(y)-1\le y(f(y)-1)-1\le -y-1,$$ i.e. $y\le -f(-1)-1$, a contradiction to the choice of $y$.

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Alternative solution.

Assume that $$f(x-f(y))\le yf(x)+x\text{ for all real }x,y.\text{\hspace{1em}(7)}$$ Let $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a)\le x$ for all real $x$ and, equivalently, $$f(y)\le y+a\text{ for all real }y.\text{\hspace{1em}(8)}$$ Now we show that $$f(z)\ge 0\text{ for all }z\ge 1\text{\hspace{1em}(9)}$$ Let $z\ge 1$ be fixed, set $b=f(z)$ and assume that $b<0$. Setting $x=w+b$ and $y=z$ in (7) gives $$f(w)-zf(w+b)\le w+b\text{ for all real }w.\text{\hspace{1em}(10)}$$ Applying (10) to $w,w+b,\dots ,w+(n-1)b$, where $n=1,2,\dots$, leads to $$\begin{array}{rl}f(w)-z^{n}f(w+nb)=& (f(w)-zf(w+b))+z(f(w+b)-zf(w+2b))\\ & +\cdots +z^{n-1}(f(w+(n-1)b)-zf(w+nb))\\ \le & (w+b)+z(w+2b)+\cdots +z^{n-1}(w+nb)\end{array}$$ From (8) we obtain $$f(w+nb)\le w+nb+a$$ and, thus, we have for all positive integers $n$ $$f(w)\le \left(1+z+\cdots +z^{n-1}+z^n\right)w+\left(1+2z+\cdots +n{z}^{n-1}+n{z}^n\right)b+z^{n}a.\text{\hspace{1em}(11)}$$ With $w=0$ we get $$a\le \left(1+2z+\cdots +n{z}^{n-1}+n{z}^n\right)b+a{z}^n\text{\hspace{1em}(12)}$$ In view of the assumption $b<0$ we find some $n$ such that $$a>(nb+a)z^n\text{\hspace{1em}(13)}$$ because the right hand side tends to $-\infty$ as $n\to \infty$. Now (12) and (13) give the desired contradiction and (9) is established. In addition, we have for $z=1$ the strict inequality $$f(1)>0.\text{\hspace{1em}(14)}$$ Indeed, assume that $f(1)=0$. Then setting $w=-1$ and $z=1$ in (11) leads to $$f(-1)\le -(n+1)+a$$ which is false if $n$ is sufficiently large. To complete the proof we set $t=\min \{-a,-2/f(1)\}$. Setting $x=1$ and $y=t$ in (7) gives $$f(1-f(t))\le tf(1)+1\le -2+1=-1\text{\hspace{1em}(15)}$$ On the other hand, by (8) and the choice of $t$ we have $f(t)\le t+a\le 0$ and hence $1-f(t)\ge 1$. The inequality (9) yields $$f(1-f(t))\ge 0$$ which contradicts (15).