To find all functions $f \colon \mathbb{R} \to \mathbb{R}$ that satisfy the inequality
$$f(y) - \left(\frac{z-y}{z-x} f(x) + \frac{y-x}{z-x} f(z)\right) \leq f\left(\frac{x+z}{2}\right) - \frac{f(x)+f(z)}{2}$$
for all real numbers $x < y < z$, we need to analyze the given condition.

First, observe that for linear functions $f(x) = ax + b$, the inequality holds with equality. This suggests that linear functions are solutions.

Next, consider quadratic functions of the form $f(x) = ax^2 + bx + c$. For these functions, the left-hand side of the inequality represents the concavity condition of the quadratic function, which is always non-positive for downward-facing parabolas (i.e., $a \leq 0$).

To confirm, we rearrange the given inequality:
$$\frac{f(z)-f(x)}{z-x} - \frac{f(y)-f\left(\frac{x+z}{2}\right)}{y-\frac{x+z}{2}} \leq \frac{f\left(\frac{x+z}{2}\right) - \frac{f(x)+f(z)}{2}}{y-\frac{x+z}{2}}.$$
This form indicates that the function $f$ must be concave and continuous, ensuring that the second derivative $f''(x) \leq 0$.

Thus, the functions that satisfy the inequality are linear functions and downward-facing parabolas. These can be expressed as:
$$f(x) = ax + b \quad \text{or} \quad f(x) = ax^2 + bx + c \quad \text{with} \quad a \leq 0.$$

The answer is: \boxed{\text{linear functions and downward-facing parabolas}}.