## $\mathrm{I}-1$

Let $\mathbb{R}$ denote the set of all real numbers. For each pair $(\alpha, \beta)$ of nonnegative real numbers subject to $\alpha+\beta \geq 2$, determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying

$$
f(x) f(y) \leq f(x y)+\alpha x+\beta y
$$

for all real numbers $x$ and $y$.


Solution. We know $f(x) f(y) \leq f(x y)+\alpha x+\beta y$ and by exchanging $x$ and $y$ we get $f(x) f(y) \leq$ $f(x y)+\beta x+\alpha y$. Combining the two we get

$$
f(x) f(y) \leq f(x y)+\gamma x+\gamma y
$$

where $\gamma=\frac{\alpha+\beta}{2}$. Notice that $\gamma \geq 1$.

Setting $x=y=-1$ in (1) we get $f(-1)^{2} \leq f(1)-2 \gamma$, so $f(1) \geq 2 \gamma$. Setting $x=y=1$ in (1) we get $f(1)^{2} \leq f(1)+2 \gamma$, so $f(1)^{2}-f(1) \leq 2 \gamma$. Since $f(1) \geq 2 \gamma \geq 2$ and $t^{2}-t$ is an increasing function for $t \geq 2$, we have $(2 \gamma)^{2}-2 \gamma \leq f(1)^{2}-f(1) \leq 2 \gamma$, hence $4 \gamma^{2} \leq 4 \gamma$, so $\gamma \leq 1$. Therefore, $\gamma=1$.

We know that $f(1) \geq 2$ and $f(1)^{2}-f(1) \leq 2$, thus necessarily $f(1)=2$. We also know $f(-1) \leq f(1)-2 \gamma=0$, so $f(-1)=0$.

Setting $x=z, y=1$ in (1) we get $2 f(z) \leq f(z)+z+1$, so $f(z) \leq z+1$. Setting $x=-z$, $y=-1$ in (1) we get $0 \leq f(z)-z-1$, so $f(z) \geq z+1$. It follows that the only function which can possibly satisfy the problem statement is

$$
f(z)=z+1
$$

It remains to check for which $\alpha$ and $\beta$ this is indeed a solution.

Substituting $f$ into original inequality, we get $(x+1)(y+1) \leq(x y+1)+\alpha x+\beta y$, thus $(1-\alpha) x+(1-\beta) y \leq 0$. This holds for all $x, y$ iff $\alpha=\beta=1$. Hence, for $(\alpha, \beta)=(1,1)$ the only solution is $f(z)=z+1$ and for $(\alpha, \beta) \neq(1,1)$ there are no solutions.

