Estonian Math Competitions

Answer: All constant functions $f(x)=c$ where $c$ is arbitrary real number.

Denote the given inequality by $V(x,y)$. Then $V(x,0)$ together with simplification gives $$f(x)\le f(0)(3)$$ for every real number $x$. On the other hand, adding $V(x,y)$ and $V(y,x)$ gives $f(x+y)\le f(xy)$, where taking $y=-x$ leads to $f(0)\le f(-x^2)$. Along with (3) this implies that $$f(z)=f(0)(4)$$ for any non-positive real number $z$. Now $V(z,-1)$ with non-positive $z$, simplified by (4), gives $f(0)\le f(-z)$. The latter along with (3) implies $f(x)=f(0)$ for all positive real numbers $x$. Thus $f(x)=f(0)$ for every real number $x$, i.e., $f$ is a constant function. All constant functions clearly satisfy the conditions of the problem.

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Alternative solution.

Denote the given inequality by $V(x,y)$. Firstly, note that $V(x,1)$ along with simplification leads to $f(x+1)\le f(1)$. As $x+1$ takes all real values, the function $f$ obtains its maximum value at 1. Secondly, note that $V(1,y)$ leads to $f(1)+f(y+1)\le 2f(y)$. Along with the inequality $f(y)\le f(1)$ obtained above, this implies $f(y+1)\le f(y)$ for all real numbers $y$. By applying the latter inequality to both $y=0$ and $y=1$ and taking into account that $f(1)$ is the maximum value of $f$, one gets $f(1)=f(0)=f(-1)$. Thirdly, note that $V(-1,y)$ gives $f(-1)+f(y-1)\le f(-y)+f(y)$. As $f(y)\le f(y-1)$ by the above, the inequality $f(-1)\le f(-y)$ must hold for every real number $y$. Since $-y$ obtains all real values, the function $f$ obtains its minimum value at $-1$. As $f(1)=f(-1)$, the maximum and minimum value coincide which means that $f$ is a constant function.