Croatian Mathematical Competitions

If we plug $y \to x - 1$ into the given inequality we get:
$$f(x) - f(x-1) \ge 1 \cdot f(1) \ge 0,$$
i.e.
$$f(x) \ge f(x-1), \quad \text{for every } x \in \mathbb{R}. \qquad (1)$$

If we plug $y \to 0$ into the given inequality we get:
$$f(x) - f(0) \ge x f(x), \qquad (2)$$

and plugging $x \to 0$, $y \to x$ gives us:
$$f(0) - f(x) \ge -x f(-x). \qquad (3)$$

Now, summing (2) and (3) gives us:
$$0 \ge x f(x) - x f(-x),$$
i.e. if we take $x > 0$
$$f(-x) \ge f(x), \quad \text{for every } x \in \mathbb{R}^{+}. \qquad (4)$$

If we plug $x \to 1$, $y \to 0$ into the given inequality we get:
$$f(1) - f(0) \ge f(1),$$
i.e.
$$f(0) \le 0. \tag{5}$$
Now we conclude that:
$$0 \stackrel{(5)}{\ge} f(0) \stackrel{(1)}{\ge} f(-1) \stackrel{(4)}{\ge} f(1) \ge 0,$$
and hence $f(-1) = f(0) = f(1) = 0$.

Repeated use of inequality (1) gives us:
$$f(x) \ge f(x-1) \ge f(x-2) \ge \dots,$$
so it follows that:
$$f(x) \ge f(x-k), \quad \text{for every } x \in \mathbb{R}, \text{ for every } k \in \mathbb{N}. \tag{6}$$

Plugging $x \to x-1$, $y \to -1$ into the given inequality gives us:
$$f(x-1) - f(-1) \ge x f(x),$$
i.e.
$$f(x-1) \ge x f(x), \quad \text{for every } x \in \mathbb{R}. \tag{7}$$

From (1) and (7) we conclude that:
$$f(x) \ge x f(x),$$
i.e.
$$f(x)(x-1) \le 0.$$

It follows from the previous inequality that
$$f(x) \le 0, \quad \text{for every } x > 1 \quad \text{and} \quad f(x) \ge 0, \quad \text{for every } x < 1. \tag{8}$$
Now we assume that $x > 1$. Then there exists $y < 1$ such that $k = x - y \in \mathbb{N}$.
Therefore:
$$0 \ge f(x) \ge f(x-k) = f(y) \ge 0$$
so we conclude that $f(x) = 0$ for every $x > 1$. Similarly, if $x < 1$ then there exists $y > 1$ such that $k = y - x \in \mathbb{N}$ so:
$$0 \ge f(y) \ge f(y-k) = f(x) \ge 0$$
and again $f(x) = 0$. Hence we conclude that the only possible solution is the function $f(x) = 0$. It is easy to check that the function $f(x) = 0$ really is a solution, i.e. that it satisfies the given conditions.