don't know how to do for all $x \leq y$ it satisfies $f(x) \leq f(y)$ means it increase in $R$ but i just found$f(x)=-2x$which not fits.

Hangzhou14Z-ZY wrote: don't know how to do for all $x \leq y$ it satisfies $f(x) \leq f(y)$ means it increase in $R$ but i just found$f(x)=-2x$which not fits. How about $f(x)\equiv0$?

qwedsazxc wrote: Hangzhou14Z-ZY wrote: don't know how to do for all $x \leq y$ it satisfies $f(x) \leq f(y)$ means it increase in $R$ but i just found$f(x)=-2x$which not fits. How about $f(x)\equiv0$? fine

Just consider the longest interval $(x_1,x_2)$.Such that $f(x_1)=f(x_2)=c$ and $c\in (x_1,x_2)$. Then try to prove $-x_1+f(-x_1)\not \in (x_1,x_2)$ or $-x_2+f(-x_2)\not \in (x_1,x_2)$. (And you need to prove such interval is exists) Then answer is $\boxed{f(x)\equiv c\quad c\in \mathbb{R}}$

Quite a difficult problem for a JBMO TST. The answer is $f \equiv c$ for any real constant $c$, which can easily be seen to work. We have the following Claim: Claim: $f(x)=0 \Longleftrightarrow f(-x)=0$, and $f(x)f(-x) \geq 0$ for all $x$. Proof: For the first part of the Claim, assume that $f(u)=0$. Then, $f(-u)=f(f(u)+u)=f(u)=0,$ as desired. For the second part, assume that a $x$ existed such that $f(x)>0>f(-x)$. Then, since $f$ is increasing, $f(-x)=f(f(x)+x) \geq f(x)>f(-x),$ a contradiction $\blacksquare$ To the problem, we distinguish two cases: Case 1: $f(x) \neq 0$ for all $x$. Then, we claim that $f$ has the same sign everywhere. Indeed, assume otherwise. Let $A<B$ be such that $f(A)f(B)<0$. Then, if $f(B)<0<f(A)$ we obtain that $f(B)<f(A) \leq f(B),$ a contradiction, while if $f(A)<0<f(B)$ we obtain that $f(-A)<0<f(-B) \leq f(-A),$ a contradiction. WLOG assume that $f(x)>0$ for all $x$. Then, $f(-x)=f(f(x)+x) \geq f(x),$ and so $f(-x) \geq f(x) \geq f(-x)$ for all $x$, that is $f(x)=f(-x)$. Now, if $A<B$ existed such that $f(A) \neq f(B)$, then since $f$ is increasing $f(A)<f(B)$, and so $f(-A)<f(-B)$, which is a contradiction since $f(-B)>f(-A) \geq f(-B)$ Thus, $f$ must be constant, as desired. The case of $f$ being everywhere negative can similarly be handled. Case 2: There exists some $x$ such that $f(x)=0$. Let $S=\{ x \in \mathbb{R} : f(x)=0 \}$. By assumption, the set $S$ is nonempty. Note that by the above Claim, if $x \in S$ then $-x \in S$. We distinguish two cases. Case 1: $S$ is upper bounded. Since $x \in S \Longleftrightarrow -x \in S$, $S$ is lower bounded, too. Note that if $x,y \in S$, then $[x,y] \subseteq S$, since $f$ is increasing. Let $a=\inf S$ and $b=\sup S$. Then, for all $\epsilon>0$ there are $x_1,x_2 \in S$ such that $a \leq x_1<a+\epsilon$ and $b-\epsilon<x_2 \leq b,$ and so $[a+\epsilon,b-\epsilon] \subseteq [x_1,x_2] \subseteq S,$ for all $\epsilon>0$, which implies that $[a,b] \subseteq S$. Since obviously $S \subseteq [a,b],$ we conclude that $S$ is an interval. Moreover, it is easy to conclude that $a=-b$. Taking a $x$ such that $x>b$ we infer that $f(-x) \leq f(a)=0=f(b) \leq f(x),$ and so by the initial Claim equality must hold, i.e. $f(x)=0$ for all $x>b$, and similarly we may extend this equality for all $x<-b$. Since for $-b \leq x \leq b$ we also have $f(x)=0$, we conclude that $f$ is zero everywhere. However, $S= \mathbb{R}$ in this case, a contradiction. Case 2: $S$ is not upper bounded. Then, for all $u \in \mathbb{R}$ there exists a $x \in S$ such that $x>\max \{-u, u \},$ and so $f(-u) \leq f(x)=0=f(-x) \leq f(u),$ that is equality must hold everywhere, which implies that $f(x)=0$ for all $x$, as desired.

My solution during the contest. Answer:$f(x)$ is constant. $f(x)=c \in R$ Let $x>0$. $i)f(x) \geq 0$ $f(x)\leq f(x+f(x))=f(-x) \leq f(x)$ $f(x)=f(-x)$ Let $x \geq y \geq -x$ $f(x) \geq f(y) \geq f(-x)=f(x)$ $ii)f(x) \leq 0$ $f(x) \leq f(x+f(x))=f(-x) \leq f(x)$ So $f(x)=f(-x)$ Again let $x \geq y \geq -x$ $f(x) \geq f(y) \geq f(-x)=f(x)$ We choose $x$ really large, so that we get $f(x)$ is constant

Easy (why the function is in JBMO TST). Suppose that there are $x,y$ such that $f(-x)$ not equal to $f(-y)$, suppose that $x \geq y$ then $f(-y) \geq f(-x)$, then $f(y+f(y)) \geq f(x+f(x))$, then $y+f(y) \geq x+f(x)$ $(1)$ (because if $x+f(x) \geq y+f(y)$ then we have $f(-x) \geq f(-y)$, which means that $f(-x)=f(-y)$). But we have $x \geq y$ and $f(x) \geq f(y)$ $(2)$. $(1)$ and $(2)$ together gives that x and y must be equal, which is contradiction to $f(-x)$ not equal to $f(-y)$. Then answer is $f(x) \equiv c$.

Nice problem, but why do such problems appear at all in JBMO TSTs?! I understand that FEs are conceptually better mathematically than almost all inequalities, but still?

I think that inequalities is more appropriate for JBMO TSTs.