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
$$f(x) \le f(0) \qquad (3)$$
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
$$f(z) = f(0) \qquad (4)$$
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.

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.