djb86 wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[f(x + y) + y \le f(f(f(x)))\] holds for all $x, y \in \mathbb{R}$. Let $P(x,y)$ be the assertion $f(x+y)+y\le f(f(f(x)))$ $P(x,f(f(x))-x)$ $\implies$ $f(f(x))\le x$ and so $P(x,y)$ implies $Q(x,y)$ : $f(x+y)+y\le f(x)$ $Q(x,y-x)$ $\implies$ $f(y)+y\le f(x)+x$ and so $\boxed{f(x)=a-x}$ which indeed is a solution whatever is $a\in\mathbb R$

pco wrote: $Q(x,y-x)$ $\implies$ $f(y)+y\le f(x)+x$ and so $\boxed{f(x)=a-x}$ which indeed is a solution whatever is $a\in\mathbb R$ That solution is very elegant, but can you explain this line? How can you conclude that fact?

Nguyenhuyhoang wrote: pco wrote: $Q(x,y-x)$ $\implies$ $f(y)+y\le f(x)+x$ and so $\boxed{f(x)=a-x}$ which indeed is a solution whatever is $a\in\mathbb R$ That solution is very elegant, but can you explain this line? How can you conclude that fact? $f(y)+y\le f(x)+x$ Swapping $x,y$, we get $f(x)+x\le f(y)+y$ And so $f(x)+x=f(y)+y$ constant $=a$ for some real $a$. And so $f(x)=a-x$

djb86 wrote: Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[f(x + y) + y \le f(f(f(x)))\] holds for all $x, y \in \mathbb{R}$. (1) $f(f(f(x)))=f(x)$: because, $P(x,0)$ $ \Rightarrow $ $f(x)\le f(f(f(x)))$ and $P(f(x), f(f(f(x)))-f(x))$ $ \Rightarrow $ $ f(f(f(x)))\le f(x) $. hence, $f(x)\le f(f(f(x)))\le f(x) $ and \[ f(f(f(x)))=f(x). \] (2) $f(x)=c-x$: ($c=f(0)$) because, from (1) we have $f(x+y)+y\le f(x) $. $P(0,x)$ $ \Rightarrow $ $ f(x)+x\le f(0)$ and $P(x, -x)$ $ \Rightarrow $ $ f(0)\le f(x)+x $. hence, $ f(x)+x=f(0) $ and \[ f(x)=c-x .\] Answer: $ f(x)=c-x $.

Easy and nice.

I am sorry Bumping old post f(f(x)) =< x imply that f(f(f(x))) =< f(x) This mean it is increasing function isn't it???? Can anyone explain me why it was increasing function???

No, in their proof, it is stated that $f(f(x)) \le x$ is true for every $x \in \mathbb{R}$, so we can change $x$ by $f(x)$.

Oh yeahhhh i am sorry too stupid argumentt

Take $y=0$ to get $f(x) \le f(f(f(x)))$ for all $x$. Take $y = f(f(x)) - x$ to get $f(f(f(x))) + f(f(x)) - x \le f(f(f(x)))$, or $f(f(x)) \le x$ for all $x$. Replacing $x$ with $f(x)$ in this inequality, we get $f(f(f(x))) \le f(x)$ for all $x$. Thus, $f(x) = f(f(f(x)))$ for all $x$. Then the given inequality becomes $f(x+y) + y \le f(x)$ for all $x$, $y$. Take $x=0$ to get $f(y) + y \le f(0)$ for all $y$, and take $y=-x$ to get $f(0) - x \le f(x) \implies f(0) \le f(x) + x$ for all $x$. Thus, $f(0) = f(x) + x$ for all $x$, so $f(x) = a-x$ where $a=f(0)$ is a constant. It is easy to check that all functions of this form satisfy the original functional inequality, so the solutions are $f(x) = a-x$ for any constant $a$. (Strikethrough edits made after reading first solution in this thread )

$(x, f(f(x))-x)$ yields $f(f(x)) \le x$. $(x, 0)$ yields $f(f(f(x))) \ge f(x)$. Combining the two gives $f(f(f(x))) = f(x)$. Thus, $f(x + y) + y \le f(x)$. Now, "swap" the roles of $x+y$ and $x$ to get $f(x) - y \le f(x + y)$. Thus, $f(x + y) = f(x) - y$. Letting $x = 0$, $f(y) = f(0) - y$. Indeed all such functions work. Yay.

Nice problem. Taking $y=f(f(x))-x$, we see $f(f(x))\le x$. Thus, the original FE can be rewritten as $$f(x+y)+y\le f^2(f(x))\le f(x)$$Swapping the sign on $y$, we see $f(x-y)-y\le f(x)$. Replacing $x$ with $x-y$, we get $f(x)+y\le f(x-y)$. It follows, $f(x)=f(x-y)-y$. Taking $x=0$, we get that $f(-y)=y+f(0)$, yielding the family of solutions $\boxed{f=-x+k}$.

$P(x,y)\implies f(x + y) + y \leqslant f(f(f(x)))$ $P(x,f(f(x))-x)\implies f(f(x))\leqslant x$ $P(0,x)\implies f(x)\leqslant f(0)-x$ $P(x,-x)\implies f(x)\geqslant f(0)-x$ $\therefore f(x)=c-x \; \text{(works)}$