function inequality $f(x+y)+y \leq f(f(f(x)))$ $f(x+y)+y \leq f(f(f(x)))$
find all possible solution for $ f: \mathbb {R} \rightarrow \mathbb {R}$

Let $y=f(f(x))-x$.  Then $f(f(f(x)))+f(f(x))-x\le f(f(f(x)))$ or $f(f(x))\le x$.  If we substitute, we obtain $f(f(f(u)))\le f(u)$.  If $y=0$, though, we know $f(x)\le f(f(f(x)))$ so we have $f(x)=f(f(f(x)))$.
Now consider $f(x)=f(f(f(x)))\ge a-x+f(x+a-x)$ and so $f(x)+x\ge f(a)+a$ for all $x,a\in\Bbb R$.  This also implies $$ f(a)+a\ge f(x)+x\ge f(a)+a\\\Longrightarrow f(x)+x=f(a)+a $$
Therefore $f(x)=f(0)-x$.  $$f(x+y)+y=f(0)-x-y+y=f(0)-x=f(x)=f(f(f(x)))$$
So this works for all $f(0)$.
