Solution. For $x=y=1$, from the statement we get $0 \leq 2 f(1) \leq f(1)$, hence $f(1)=0$.

For $y=1$, the relation from the statement leads to $x \lg x \leq f(x)$, for any $x>0$.

For $y=\frac{1}{x}$, from the statement we obtain $0 \leq \frac{f(x)}{x}+x f\left(\frac{1}{x}\right) \leq 0$, so $f\left(\frac{1}{x}\right)=-\frac{f(x)}{x^{2}}$, for any $x>0$.

For $y=\frac{1}{x}$, from (1) we get $\frac{1}{x} \lg \left(\frac{1}{x}\right) \leq f\left(\frac{1}{x}\right)$, so $f\left(\frac{1}{x}\right) \geq-\frac{\lg x}{x}$ and, taking into account (2), we obtain

$$
-\frac{f(x)}{x^{2}} \geq-\frac{\lg x}{x} \Leftrightarrow f(x) \leq x \lg x, \text { for any } x>0
$$

From (1) and (3) it follows that $f(x)=x \lg x$, for any $x>0$, which is a function that satisfies the condition from the statement.
Problem 3. Determine the functions that satisfy the conditions

$$
x y \lg (x y) \leq y f(x)+x f(y) \leq f(x y) \text{, for all } x, y>0 \text{. }
$$