TITLE: $f(x+y) + f( f(x) + f(y) ) = f( f( x+f(y) ) + f( y+f(x) ) )$
QUESTION [9 upvotes]: Suppose $f\colon \mathbb R\to\mathbb R$ is a strictly decreasing function which satisfies the relation 
 $$f(x+y) + f( f(x) + f(y) ) = f( f( x+f(y) ) + f( y+f(x) ) ) , \quad \forall x , y \in\mathbb R $$
Find a relation between $f( f(x))$ and $x$.

REPLY [13 votes]: I think I have got hold of something , putting y=x in the functional equation we get 
$$
f(2x) + f(2f(x)) = f(2f(x+f(x)))
$$
Changing $x$ to $f(x)$ we also get 
$$
f(2f(x)) + f(2f(f(x))) = f(2f(f(x)+f(f(x))))
$$
Subtracting the former from the later equation we get 
$$
f(2f(f(x))) - f(2x) = f(2f(f(x)+f(f(x)))) - f(2f(x+f(x)))
$$
Now since $f(x)$ is strictly decreasing $x>y$ if and only if $f(x) < f(y)$ . Assume that $f(f(x)) > x$ , for some $x$, then 
$$
\begin{align}
2f(f(x)) > 2x&\Longleftrightarrow f(2f(f(x))) < f(2x)\\
&\Longleftrightarrow f(2f(f(x)+f(f(x))))< f(2f(x+f(x)))\\
&\Longleftrightarrow 2f(f(x)+f(f(x))) > 2f(x+f(x))\\
&\Longleftrightarrow f(f(x)+f(f(x))) > f(x+f(x))\\
&\Longleftrightarrow f(x)+f(f(x)) < x+f(x)\\
&\Longleftrightarrow f(f(x)) < x
\end{align}
$$
contradiction! So  , $f(f(x)) = x$ , for all real $x$.