i think you are kidding ,just set $x=y$

Muradjl wrote: i think you are kidding ,just set $x=y$ He's right, a small discussion proves injectivity and the result follows.

steppewolf wrote: Determine all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that:$$f(\max \left\{ x, y \right\} + \min \left\{ f(x), f(y) \right\}) = x+y $$for all real $x,y \in \mathbf{R}$ Let $P(x,y)$ be the assertion $$f(\max \left\{ x, y \right\} + \min \left\{ f(x), f(y) \right\}) = x+y $$$P(x,x)$ $\implies$ $f(x+f(x))=2x$ for all real $x$. $P(x,y)$ with $x>y$ $\implies$ $$f(x+ \min \left\{ f(x), f(y) \right\}) = x+y $$. If $f(y)\geq f(x)$ then $$f(x+f(x))=2x=x+y$$$\implies$ $x=y$, but $x>y$. So if $x>y$ then $f(x)>f(y)$.And we find that if $x>y$ Then $f(x+f(y))=x+y$ $\implies$ $f(x)=x$ for all real number $x$.

Mathuzb wrote: Then $f(x+f(y))=x+y$ $\implies$ $f(x)=x$ for all real number $x$. Why this implication ?

If I am correct, this is why: If $f(y)=a$, put $y=0$, you get $f(x+a)=x$, so $f(x)=x-a$, putting $x=0$, we get $a=0-a$, so $a=0$, thus $f(x)=x$

steppewolf wrote: Determine all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that:$$f(\max \left\{ x, y \right\} + \min \left\{ f(x), f(y) \right\}) = x+y $$for all real $x,y \in \mathbf{R}$ Let $P(x,y)$ be the assertion $f(\max(x,y)+\min(f(x),f(y)))=x+y$ Let $a=f(0)$ $P(x,x)$ $\implies$ $f(x+f(x))=2x$ Let $x>y$. If $f(x)\le f(y)$, then $P(x,y)$ $\implies$ $f(x+f(x))=x+y=2x$ and so $x=y$, impossible. So $x>y$ $\implies$ $f(x)>f(y)$ and $P(x,y$ $\implies$ : $f(x+f(y))=x+y$ $\forall x>y$ (I) So $f(x)=x+y-f(y)$ $\forall x>f(y)$ Setting there $y=0$, we get $f(x)=x-a$ $\forall x>a$ (II) Let then $y\in\mathbb R$ and $x>\max(y,a-f(y))$ : $x>y$ implies $f(x+f(y))=x+y$ (see I above) $x+f(y)>a$ implies $f(x+f(y))=x+f(y)-a$ (see II above) And so $f(y)=y+a$ $\forall y$ Plugging this back in original equation, we get $a=0$ and so $\boxed{f(x)=x\quad\forall x}$

pco wrote: Mathuzb wrote: Then $f(x+f(y))=x+y$ $\implies$ $f(x)=x$ for all real number $x$. Why this implication ? We have for all $x>y$ $\implies$ $f(x+f(y))=x+y$. Then $x>0$ $\implies$ $f(x+f(0))=x$ and $f(0)=0$. So $f(x)=x$ for positive real. Similarly we find $f(x)=x$ for all negative real.

Let me answer @pco question to @Mathuzb sol: Bc that implies $f$ is linear (we'll show $f(y)-y=f(z)-z$ for all $y$ with fixed $z$, consider $t>\max \{ y+f(y),z+f(z)\}$ and substitute $x$ by $t-f(y)$ and $t-f(z)$.) Then let $f(x)=x+c$ for all real $x$ for a constant $c$, and so $$x+y=f(\max \left\{ x, y \right\} + \min \left\{ f(x), f(y) \right\}) =f(\max \left\{ x, y \right\} + \min \left\{ x, y \right\} +c)=f(x+y+c)=x+y+2c\implies c=0.$$

This was my proposal

Let $P(x,y)$ be the given assertion. Clearly $f$ is surjective. So there exists a real number $u$ such that $f(u)=0$. $P(u,u)\implies u=0$, so $f(x)=0\iff x=0$. $P(x,x)\implies f(x+f(x))=2x$ Suppose there exists $a>b$ such that $f(a)\leq f(b)$. $P(a,b)\implies 2a=f(a+f(a))=a+b$, contradiction. Hence $f$ is strictly increasing. Suppose $x>0$. Then $P(x,0)\implies f(x)=x$ $P(x,-x)\implies f(x+f(-x))=0\implies f(-x)=-x$ Thus $f(x)=x$ for all $x\in\mathbb R$, which indeed is a solution.

Pretty cute! The proof is structured in several Claims. Claim 1: $f(0)=0$. Proof: Take $x=y$ to obtain $f(x+f(x))=2x$. Thus, $f$ is surjective. However, if $f(u)=0$ then $2u=f(u+f(u))=f(u)=0$, and so $u=0$. Hence, $f(0)=0$, as desired $\blacksquare$ Claim 2: $f(x)$ and $x$ always have the same sign. Proof: Suppose firstly that $x <0$. Then, for $y=0$ in the given equation, $f(\min \{f(x),0 \})=x$. Therefore, if $f(x)>0$ then we would have $f(0)=x,$ that is $x=0$, a contradiction. Hence, $f(x)<0$ for all $x<0$. Now, if $x>0$, we take again $y=0$ in the given equation to obtain $f(x+\min \{f(x),0 \})=x$. Therefore, if $f(x)<0$ then we would have $x=f(x+f(x))=2x,$ and so $x=0$, a contradiction. Therefore, $f(x)>0$ for all $x>0$ $\blacksquare$ Claim 3: $f(x)=x$ for all $x \geq 0$. Proof: We may take $x \geq 0$ and $y=0$ in the given equation and use Claim 2 $\blacksquare$ Claim 4: $f(x)=x$ for all $x \leq 0$. Proof: Take $y \geq 0 \geq x$ such that $y>-f(x),$ too. Then $x+y=f(\max \left\{ x, y \right\} + \min \left\{ f(x), f(y) \right\})=f(y+f(x))=y+f(x),$ and so $f(x)=x,$ as desired $\blacksquare$ To finish, it is easy to check that the identity function works.