We claim that only $\boxed{f(x) \equiv x}$ satisfies the conditions. It is easy to check that it satisfies the conditions. From $(1)$, $f$ is unbounded. Let $P(x, y): f(xy) \ge f(x)f(y)$. $P(x, x): f(x^2) \ge f(x)^2 \ge 0 \Leftrightarrow f(x) \ge 0$ for all $x \ge 0$. $P(0, x): f(0) \ge f(0)f(x) \Leftrightarrow f(0)(1 - f(x)) \ge 0$. for all $x$. Since $f$ is unbounded, there exists $x$ such that $f(x) > 1$. Together with $f(0) \ge 0$, we have $f(0) = 0$. $P(1, 1): f(1) \ge f(1)^2 \Leftrightarrow f(1)(f(1) - 1) \le 0$. Since $f(1) \ge f(0) + 1 = 1$, we have $f(1) = 1$. $P(-1, -1): f(1) \ge f(-1)^2 \Leftrightarrow f(-1)^2 \le 1 \Leftrightarrow -1 \le f(-1) \le 1$. Since $f(-1) \le f(0) - 1 = -1$, we have $f(-1) = -1$. $P(x, -1): f(-x) \ge -f(x)$ for all $x \in \mathbb{R}$. By $(1)$, $f(x + n) \ge f(x) + n$ for all $x \in \mathbb{R}$ and $n \in \mathbb{N}$. By $(2)$, $f(x^n) \ge f(x)^n$ for all $x > 0$ and $n \in \mathbb{N}$. Note that $f(x) = f(\lfloor x \rfloor + \{ x \}) \ge f(\{ x \}) + \lfloor x \rfloor \ge \lfloor x \rfloor > x - 1$ for all $x \ge 0$. So for all $x > 1$, $(x - 1)f\left(\dfrac{1}{x}\right) < f(x)f\left(\dfrac{1}{x}\right) \le 1 \Leftrightarrow f\left(\dfrac{1}{x}\right) < \dfrac{1}{x - 1}.$ Plugging $x \to \dfrac{1}{x}$, we get $f(x) < \dfrac{x}{1 - x}$ for all $x \in (0, 1)$. Then $f(x)^n \le f(x^n) < \dfrac{x^n}{1 - x^n} \Leftrightarrow f(x) < \dfrac{x}{\sqrt[n]{1 - x^n}}$ for all $x \in (0, 1)$. As $n$ goes to infinity, $\sqrt[n]{1 - x^n}$ approaches 1. Thus $f(x) \le x$ for all $x \in (0, 1)$. $f(-x) \ge -f(x)$ and $-f(x) \ge -x$ for all $x \in (0, 1)$ is equivalent to $f(x) \ge x$ for all $x \in (-1, 0)$. For all $x \in (-1, 0)$, $x = (x + 1) - 1 \ge f(x + 1) - 1 \ge f(x) \ge x$. Since all equalities have to hold, $f(x) = x$ and $f(x + 1) = x + 1$ which means $f(x) = x$ for all $x \in [-1, 1]$. For all $x \neq 0$ there exists $c \in (0, 1)$ such that $|cx| < 1$. $P(x, c): cx \ge cf(x) \Leftrightarrow f(x) \le x$. $P(x, -c): -cx \ge -cf(x) \Leftrightarrow f(x) \ge x$. Thus, $f(x) = x$ for all $x \neq 0$, and with $f(0) = 0$ we conclude that $f(x) \equiv x$ is the only solution.