Problem 3. Determine the functions $f:(0, \infty) \rightarrow \mathbb{R}$ with the property

$$
\ln (x y) \leq f(x)+f(y)-x-y \leq f(x y)-x y, \quad(\forall) x, y \in(0, \infty)
$$
Problem 3. Determine the functions $f:(0, \infty) \rightarrow \mathbb{R}$ with the property

$$
\ln (x y) \leq f(x)+f(y)-x-y \leq f(x y)-x y, \quad(\forall) x, y \in(0, \infty)
$$

## Grading Rubric.

(2p) From $\ln (x y) \leq f(x y)-x y$, it follows that $f(t)-t \geq \ln t,(\forall) t \in(0, \infty)$.

(2p) If we denote $f(x)-x=g(x)$, then

$$
\ln (x y) \leq g(x)+g(y) \leq g(x y), \quad(\forall) x, y \in(0, \infty)
$$

For $x=y=1$ we have $0 \leq 2 g(1) \leq g(1)$, i.e., $g(1)=0$.

(3p) For $y=\frac{1}{x}$, we obtain:

$$
0 \leq g(x)+g\left(\frac{1}{x}\right) \leq 0 \Rightarrow g(x)+g\left(\frac{1}{x}\right)=0 \Rightarrow(g(x)-\ln x)+\left(g\left(\frac{1}{x}\right)-\ln \frac{1}{x}\right)=0
$$

Since $g(t) \geq \ln t$, for any $t \in(0, \infty)$, it follows that $g(x)-\ln x=g\left(\frac{1}{x}\right)-\ln \frac{1}{x}=0$, i.e., $f(x)=\ln x+x$.