Problem 1358

1. Let $P(x, y, z)$ denote the assertion of the given functional inequality:
$$\frac{f(xy) + f(xz)}{2} - f(x)f(yz) \geq \frac{1}{4}$$
for all $x, y, z \in \mathbb{R}$.

2. Consider $P(1, 1, z)$:
$$\frac{f(1 \cdot 1) + f(1 \cdot z)}{2} - f(1)f(1 \cdot z) \geq \frac{1}{4}$$
Simplifying, we get:
$$\frac{f(1) + f(z)}{2} - f(1)f(z) \geq \frac{1}{4}$$
Let $f(1) = a$. Then:
$$\frac{a + f(z)}{2} - af(z) \geq \frac{1}{4}$$
Rearranging, we obtain:
$$\frac{a + f(z)}{2} - af(z) \geq \frac{1}{4}$$
$$a + f(z) - 2af(z) \geq \frac{1}{2}$$
$$a + f(z)(1 - 2a) \geq \frac{1}{2}$$
$$f(z)(1 - 2a) \geq \frac{1}{2} - a$$

3. Consider $P(x, 1, 1)$:
$$\frac{f(x \cdot 1) + f(x \cdot 1)}{2} - f(x)f(1 \cdot 1) \geq \frac{1}{4}$$
Simplifying, we get:
$$f(x) - f(x)f(1) \geq \frac{1}{4}$$
Let $f(1) = a$. Then:
$$f(x) - af(x) \geq \frac{1}{4}$$
$$f(x)(1 - a) \geq \frac{1}{4}$$
Since $f(x) \geq \frac{1}{4(1 - a)}$, we need $1 - a \neq 0$, i.e., $a \neq 1$.

4. Consider $P(x, 0, 0)$:
$$\frac{f(x \cdot 0) + f(x \cdot 0)}{2} - f(x)f(0 \cdot 0) \geq \frac{1}{4}$$
Simplifying, we get:
$$f(0) - f(x)f(0) \geq \frac{1}{4}$$
Let $f(0) = b$. Then:
$$b - bf(x) \geq \frac{1}{4}$$
$$b(1 - f(x)) \geq \frac{1}{4}$$
Since $b \geq \frac{1}{4(1 - f(x))}$, we need $1 - f(x) \neq 0$, i.e., $f(x) \neq 1$.

5. From the previous steps, we have:
$$f(x) \geq \frac{1}{2} \quad \text{and} \quad f(x) \leq \frac{1}{2}$$
Therefore, $f(x) = \frac{1}{2}$ for all $x \in \mathbb{R}$.

6. Finally, we verify that $f(x) = \frac{1}{2}$ satisfies the original inequality:
$$\frac{f(xy) + f(xz)}{2} - f(x)f(yz) = \frac{\frac{1}{2} + \frac{1}{2}}{2} - \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$$
which is indeed $\geq \frac{1}{4}$.

Thus, the only function that satisfies the given inequality is $f(x) = \frac{1}{2}$ for all $x \in \mathbb{R}$.

The final answer is $\boxed{f(x) = \frac{1}{2} \text{ for all } x \in \mathbb{R}}$.