The solutions to the given functional inequality are $f(x) = 0$ for all $x$ and $f(x) = x$ for all $x$. For $f(x) = 0$, we easily find that equality always holds. For $f(x) = x$ we check that
$$\begin{aligned} (a-b)(c-d) + (a-d)(b-c) &= ac - ad - bc + bd + ab - ac - bd + cd \\ &= -ad - bc + ab + cd \\ &= (a-c)(b-d), \end{aligned}$$
so equality holds in that case too. Now we show that these functions are the only two solutions.

Substituting $a = b = c = d = 0$ gives us $2f(0)^2 \le 0$, and therefore $f(0) = 0$.
Then we substitute $b = a - x$, $c = a$, and $d = a - y$, which gives $a - b = x$, $a - c = 0$ and $a - d = y$. From this we deduce that
$$f(y)(f(x) + f(-x)) \le 0 \qquad (1)$$
Suppose there is a $y$ such that $f(y) \ne 0$. If we then substitute $x = y$ in the equation above and move one of the terms to the right, we find that
$$0 < f(y)^2 \le -f(y)f(-y).$$
Therefore one of the two values $f(y)$ and $f(-y)$ is positive and the other is negative. Assume without loss of generality that $f(y)$ is positive.
Now, given arbitrary $a$ and $y$, substitute $b = a$, $c = 0$, and $d = a - y$. Then we find that $f(y)f(a) \le af(y)$. Thus, if we divide both sides by the positive $f(y)$, we get $f(a) \le a$. On the other hand, if we substitute $b = a$, $c = 0$, and $d = a + y$, we find that $f(-y)f(a) \le af(-y)$. Since $f(-y)$ is negative, the sign flips when we divide by $f(-y)$ so we deduce that $f(a) \ge a$. We conclude that $f(a) = a$ for all real $a$.
Therefore a solution $f$ of the given functional inequality is either the zero function or $f(a) = a$ for all real $a$, and we confirmed in the beginning that both are indeed solutions. $\square$