5. Find all functions f defined on the set of real numbers such that $\mathrm{f}(\mathrm{xy})+\mathrm{f}(\mathrm{xz}) \geq 1+\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{yz})$ for all real $\mathrm{x}, \mathrm{y}, \mathrm{z}$.
Answer: $\mathrm{f}(\mathrm{x})=1$ (a constant function, all values of which are equal to 1).

Instructions. Let $\mathrm{x}=\mathrm{y}=\mathrm{z}=1$ and we get $\mathrm{f}(1)+\mathrm{f}(1) \geq 1+\mathrm{f}(1) \mathrm{f}(1)$, or $(f(1)-1)^{2} \leq 0$ and thus, $f(1)=1$.

Let $\mathrm{x}=\mathrm{y}=\mathrm{z}=0$ and we get $\mathrm{f}(0)+\mathrm{f}(0) \geq 1+\mathrm{f}(0) \mathrm{f}(0)$, or $(\mathrm{f}(0)-1)^{2} \leq 0$ and thus, $\mathrm{f}(0)=1$.

Let $\mathrm{z}=\mathrm{y}=1$ and we get $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}) \geq 1+\mathrm{f}(\mathrm{x}) \mathrm{f}(1)$, from which $\mathrm{f}(\mathrm{x}) \geq 1$.

Let $\mathrm{z}=\mathrm{y}=0$ and we get $\mathrm{f}(0)+\mathrm{f}(0) \geq 1+\mathrm{f}(\mathrm{x}) \mathrm{f}(0)$, from which $\mathrm{f}(\mathrm{x}) \leq 1$.

Therefore, $\mathrm{f}(\mathrm{x})=1$.

Verification shows that this function fits.

Criteria. The value of the function at zero and one is found: 2 points.