Not even an hour and a half past since FKMO Day 1 ended, and all three problems has been posted. Nice.

for $x$, denote the set of $y$ satisfing the condition as $A(x)$ and the set of $y$ satisfing $y\le x$ , $f(y)\le f(x)$ as $E(x)$ Claim) $E(x)$ is a finite set.

lets take any $a,b$ ($a<b$) and prove that $f(a)>f(b)$. if $a\notin E(b)$, it is trivial. if $a\in E(b)$, as $E(b)$ is a finite set, we can find a smallest element of $E(b)$ that is bigger than $a$, and denote it as $c$. If such element doesnt exist, let $c=b$. $\therefore$ for any $x\in (a,c), x\notin E(b)$ $\therefore$ for any $x\in (a,c), f(x)>f(b)\ge f(a)$ lets take any element of $A(x)$, and denote it as $g(x)$. $(a+f(g(x)))(g(x)+f(a))\le (x+f(g(x)))(g(x)+f(x))\le 4$ $\therefore g(x)\in A(a)$ since $A(a)$ is a finite set and there is infinite $x$ in $(a,c)$, $\therefore$ there exists $k$ such that there is infinite $x$ such that $k=g(x)$ $\therefore$ $A(k)$ is a infinite set contradiction.

This looks like an IMO 2022/2 reboot. We prove something slightly stronger: Quote: Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function such that for any positivr eal number $x{}$ there exists at least one, but countably many positive real numbers $y{}$ for which \[(x+f(y))(y+f(x))\leqslant 4.\]Prove that $f$ is strictly decreasing. Denote the given condition by $(\dagger)$. For any $x{}$ let $S_x$ be the set of positive real numbers $y{}$ for which $(\dagger)$ is satisfied. Notice that $x\in S_y$ if and only if $y\in S_x$. We know that $S_x$ is non-empty for all $x{}$. Hence, for any $x{}$ there exists $y{}$ such that $x\in S_y$. Now, we prove the following: Claim: Suppose that $a<b$ and for some $x{}$ we have $a\notin S_x$ but $b\in S_x$. Then, $f(a)>f(b)$. Proof: Simply note that from $(\dagger)$ applied for the pairs $(x,a)$ and $(x,b)$ we get \[\frac{4}{x+f(a)}<f(x)+a<f(x)+b\leqslant \frac{4}{x+f(b)},\]hence $f(a)>f(b)$ follows. $\square$ Now, for each $x{}$ define the set $S_x^<=\{y:y<x, f(y)\leqslant f(x)\}$. Choose $y{}$ such that $x\in S_y$. Then, the claim implies that $S_x^<\subseteq S_y$, which is countable. Suppose that $S_x^<$ is non-empty. Choose $y\in S_x^<$ and $z\in (y,x)\setminus S_x^<$ (which exists, as $S_x^<$ is countable). Then, $y<z<x$ and $f(y)\leqslant f(x)<f(z)$. Then, \[(y+f(t))(t+f(y))\leqslant (z+f(t))(t+f(z)),\]so $z\in S_t$ implies $t\in S_z$ which implies $t\in S_y$. As $z\in (y,x)\setminus S_x^<$, there are uncountably many choices for $z{}$ and because each $S_t$ is countable, we get uncountably many values of $t{}$. That being said, $S_y$ is uncountable, which is a contradiction.

Call $x$ and $y$ pals if $(x+f(y))(y+f(x)) \leq 4$; note that this is obviously a symmetric relation. Suppose that we had some $a<b$ with $f(a) \leq f(b)$. Consider the interval $(a,b)$: if some $c \in (a,b)$ has $f(c)<f(b)$, then any pals of $b$ are also pals of $c$. Hence finitely many $c$ must exist, else the pals of $b$ have infinitely many pals. Therefore we can find an infinite set $S \subseteq (a,b)$ such that for any $c \in S$, we have $f(c) \geq f(b)$. Consider the pals of the elements of $S$. If finitely many reals can possibly be pals, then by infinite pigeonhole one of these pals must have infinitely many pals: contradiction. However, any pal of an element in $S$ is also a pal of $a$, so if we have an infinite number of reals that are pals of some element in $S$, then $a$ has infinitely many pals: also a contradiction. $\blacksquare$

This is similar which appeared above, however, with more systemetic definition of notations for comprehension. * * * Suppose that the function $f$ satisfies the condition. For positive reals $x$, $y$, define $x \sim y \iff (x+f(y))(y+f(x))$. Note that $\sim $ is symmetric. Let $[x]$ denote $\left\{y>0| x \sim y\right\}$. For a positive $a$, define $x \preceq a$ by $x \le a \wedge f(x) \le f(a)$. Note that $\preceq$ is reflexive and transitive. More importantly, we have $$ a \preceq b \wedge b \sim c \implies a \sim c $$ Now define $P_a = \left\{ x>0| x \preceq a \right\}$. Let $p \sim a$. If $x \in P_a$, then $x \preceq a$ so $x \sim p$. Hene $P_a \subseteq [p]$ so $P_a$ is finite. Now define $S_a = \left\{ x>0| a \preceq x \right\}$. $S_a$ should be finite. To see this, suppose that $S_a$ is infinite. For each $x \in S_a$, we have $[x] \subseteq [a]$, $$ \bigcup_{x \in S_a} [x] \subseteq [a] $$Therefore, there exists $q \in [a]$ such that $q \in [x]$ for infinitely many $x \in S_a$. But this implies that $[q]$ is infinite which is an absurdity. Now, we are prepared. Suppose that $a<b$ whereas $f(a) \le f(b)$. Since $a \preceq b$ and $P_b$ is finite, we have infinite set $N=(a,b) \setminus P_b$. But then, $ x \in N$ implies $f(x) > f(b)$ and so $a \preceq x$ that $N \subseteq S_a$ which contradicts to that finiteness of $S_a$.

let $(f(x)+y)(f(y)+x)\leq 4 \rightarrow y \in S_x$ claim) if $a<b$, $a\notin S_x, b\in S_x \rightarrow f(a)>f(b)$ proof) $(f(a)+x)(f(x)+a)>4\geq (f(b)+x)(f(x)+b)>(f(b)+x)(f(x)+a)$, therefore $f(a)+x>f(b)+x, f(a)>f(b)$ $a\in S_x\rightarrow x\in S_a$, which is trivial. let's show $a<b, f(a)\leq f(b)\rightarrow S_b \in S_a$ $a+f(t)<b+f(t), f(a)+t\leq f(b)+t \rightarrow (a+f(t))(f(a)+t)<(b+f(t))(f(b)+t),$ therefore $(b+f(t))(f(b)+t)\leq 4\rightarrow (a+f(t))(f(a)+t)\leq 4$ done. Now let's show there is a contradiction. First, let $a<c<b$, $c\notin S_x$ and for every $c, f(a)\leq$ (by the claim)$f(b)<f(c) \rightarrow f(a)<f(c),$ which means that if $t\in S_c\rightarrow t\in S_a$. Since $S_x$ is countable, the boundary of $c$ is uncountable. And because the boundary of $c$ is uncountable, $c\in S_t$, therefore there are uncountably many $t$ which is possible. And $t\in S_a$, therefore $S_a$ is uncountable, and this is contradiction.