The 36th KOREAN MATHEMATICAL OLYMPIAD Final Round

For each positive real number $x$, let $A_x$ be the set of positive real numbers $y$ satisfying $(x+f(y))(y+f(x))\le 4$. The following are easy consequences by the definition. (1) If $x\in A_y$, then $y\in A_x$. (2) For $x<y$, if $f(x)\le f(y)$ then $A_y\subseteq A_x$. We prove the following lemma. Lemma. For $x\in {\mathbb{R}}^{+}$, there are only finitely many positive real numbers $y(<x)$ such that $f(y)<f(x)$. Proof. Since $A_x$ is not empty, there exists $z\in A_x$. For $y(<x)$ with $f(y)<f(x)$, we have $A_x\subseteq A_y$ by (2). So $z$ also belongs to $A_y$, and $y\in A_z$ by (1). Therefore, if there are infinitely many $y(<x)$ such that $f(y)<f(x)$, $A_z$ becomes an infinite set, which is a contradiction. $\square$ Assume that there exist $a,b\in {\mathbb{R}}^{+}$ such that $a<b$ and $f(a)\le f(b)$. Let $S=\{x\mid a<x<b,f(x)>f(a)\}$. By the lemma, there are only finitely many $x$ with $a<x<b$ such that $f(x)<f(b)$. So $S$ is an infinite set. For any $s\in S$, $A_s\subseteq A_a$ by (2). Since $A_s$ is not empty, there exists $t\in A_s\subseteq A_a$. So, $s\in A_t\subseteq {\bigcup }_{z\in A_a}{A}_z$ by (1). That is, $S\subseteq {\bigcup }_{z\in A_a}{A}_z$ which yields a contradiction because $S$ is an infinite set but $$|\bigcup _{z\in A_a}{A}_z|\le \sum _{z\in A_a}|A_z|$$ is finite. $\square$