Quidditch wrote: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $$f(\max \left\{ x, y \right\} + \min \left\{ f(x), f(y) \right\}) = x+y $$for all $x,y \in \mathbb{R}$. Let $P(x,y)$ be the assertion $f(\max(x,y)+\min(f(x),f(y)))=x+y$ $f(x)$ is surjective. If $f(u)=f(v)$ for some $u,v$, let $y>\max(u,v)$ : Comparing $P(u,y)$ with $P(v,y)$, we get $u=v$ and so $f(x)$ is injective. Let $u$ such that $f(u)=0$ : $P(u,u)$ $\implies$ $u=0$ and so $f(0)=0$ Let $x>y$. If $f(x)\le f(y)$ : $P(x,y)$ $\implies$ $f(x+f(x))=x+y$ but $P(x,x)$ $\implies$ $f(x+f(x))=2x$ and so contradiction. So $x>y$ $\implies$ $f(x)>f(y)$ and $f(x)$ is strictly increasing. Let $x>0$ (so that $f(x)>0$) : $P(x,0)$ $\implies$ $f(x)=x$ $\forall x>0$ Let $x<0$ (so that $f(x)<0$) and $y>-x$ (so that $x+y>0$ and so $f(x+y)=x+y$) : $P(y,x)$ $\implies$ $f(y+f(x))=y+x=f(y+x)$ and so, since injective, $f(x)=x$ $\forall x<0$ Hence the answer $\boxed{f(x)=x\quad\forall x}$, which indeed fits.

https://aops.com/community/p23123029 Let $P(x,y)$ be the assertion $f(\max\{x,y\}+\min\{f(x),f(y)\})=x+y$. $P(0,0)\Rightarrow f(f(0))=0$ $P(f(0),f(0))\Rightarrow f(0)=0$ For positive $x$, if $f(x)>0$ then $P(x,0)\Rightarrow f(x)=x$ whereas if $f(x)\le0$ then $P(x,0)\Rightarrow f(x+f(x))=x$, a contradiction by $P(x,x)$. Then $f(x)=x$ for all $x>0$. For negative $x$, if $f(x)\ge0$ then $P(x,0)\Rightarrow x=0$, so $f(x)<0$. Then: $P(x,0)\Rightarrow f(f(x))=x$ $P(x,x)\Rightarrow f(x+f(x))=2x$ $P(f(x),f(x))\Rightarrow f(x)=x$ In conclusion, $\boxed{f(x)=x}$ for all $x$, which fits.

$P(0,x)\implies f$ is surjective $P(0,0) \implies f(f(0)) = 0$ $P(f(0),f(0))\implies f(0) = 0$ For $x > 0$ such that $f(x)\ge 0$, $P(x,0)\implies f(x) = x$ For $x < 0$ such that $f(x)\ge 0$, $P(x,0)\implies f(0) = x$, a contradiction It follows that if $f(x) = c$ and $c > 0$, then $x = c$ For $x > 0$ such that $f(x)\le 0$, $P(x,0)\implies f(x + f(x)) = x\implies f(x) = 0$, a contradiction because $f(x + f(x)) = x$ Hence $f(x) = x$ for all $x\ge 0$ and $f(f(x)) = x$ for all $x$ Comparing $P(x,x)$ and $P(f(x),f(x))\implies f(x)\equiv x$ which is a solution

Obvously we have that $f$ must be surjective. Assume there exists a $y$ s.t. $x \geq y$ and $f(x) \leq f(y)$, then we have that $P(x,y)$ turns into: $$f(x+f(x))=x+y$$Looking at $P(x,x)$ we have that $f(x+f(x))=2x$, this implies that $x=y$ Thus if $x > y \implies f(x) > f(y)$. Thus set $y$ to be less than $x$, we have that: $$ f(x+f(y))=x+y $$obviously this is injective (choose a $x$ such that $x > y_1 >y_2$, where $f(y_1)=f(y_2)$). Now set $x+y > 0$, thus we have that: $$f(x+f(y))=x+y=f(x+y+f(0))$$thus $x+f(y)=x+y+f(0)$, plugging in what we have we get that $f(x)=x$ is the only solution

This is the same as Macedonian MO 2018/3