Hard functional equation: $ f \big ( x y + f ( x ) \big) = f \big( f ( x ) f ( y ) \big) + x $ 
Let $ \mathbb R _ { > 0 } $ be the set of positive real numbers. Find all functions $ f : \mathbb R _ { > 0 } \to \mathbb R _ { > 0 } $ such that
$$ f \big ( x y + f ( x ) \big) = f \big( f ( x ) f ( y ) \big) + x $$
for all positive real numbers $ x $ and $ y $.

What I thought: We could change $ x $ by $ y $, and then subtract.
Source: Brazil National Olympiad 2019 #3
 A: Exchanging $x$ and $y$ and substracting, it follows $f(xy+f(x))-f(xy+f(y))=x-y$. In particular, if $f(x)=f(y)$ then $x=y$. 
The equation also tells us that if $r > f(x)$, we can find a $y> 0$ such that $r=f(x)+xy$, so $f(r)=f(xy+f(x))=f(f(x)f(y))+x > x$, ie that if $r > f(x)$, $f(r) > x$. 
In particular, if $x > f(x)$, $f(x) > x$, so we have, for all $x$, $f(x) \geq x$. 
Now, let us fix some $x > 0$ such that $f(x)>x$. 
Define, for any $y > 0$, $g(y)=\frac{f(x)}{x}(f(y)-1)$. If $g(y)>0$, then note that $xg(y)+f(x)=f(x)f(y)$, thus $f(xy+f(x))=f(xg(y)+f(x))+x$. 
Therefore, if $y >0$ and $g^n(y)>0$ is defined, $0<f(xg^n(y)+f(x))=f(xy+f(x))-nx$. As a consequence, $n < \frac{f(xy+f(x))}{x}+1$ (the precise estimate is irrelevant, just remember tha the RHS is explicit in $x$ and $y$). 
In particular, there exists some $n \geq 0$ (depending on $x,y$) such that  $g^n(y) > 0$ is defined and $g^{n+1}(y) \leq 0$. 
Now, take $y > \alpha$, where $f(x)(\alpha-1)=x\alpha$. Then $g(y)=\frac{f(x)}{x}(f(y)-1) \geq \frac{f(x)}{x}(y-1) > f(x)(\alpha-1)/x=\alpha$. 
We find that $g^n(y)$ is defined and positive for all $n$, a contradiction. 
A: $$f(xy + f(x)) = f(f(x)\cdot f(y)) + x.$$
Let us substitute $y = 1$:
$$f(x + f(x)) = f(f(x)\cdot f(1)) + x.$$
Now, let us substitute into the initial equation $x = 1$:
$$f(y + f(1)) = f(f(1)\cdot f(y)) + 1.$$
In the latter equation, let us replace $y$ by $x$:
$$f(x + f(1)) = f(f(x)\cdot f(1)) + 1.$$
Now, we have 
$$
f(x + f(x)) = f(f(x)\cdot f(1)) + x \\
f(x + f(1)) = f(f(x)\cdot f(1)) + 1
$$
Let $g(x) = f(f(x)\cdot f(1))$. Then, we have
$$
f(x + f(x)) = g(x) + x \\
f(x + f(1)) = g(x) + 1
$$
We see that a linear shift in the argument of the function $f(x)$ results in a linear shift in the values of function g(x).


*

*Shifting by $f(x)$, i.e. $f(x + f(x))$ implies shifting by $x$.

*Shifting by $f(1)$, i.e. $f(x + f(1))$ implies shifting by $1$.  


This is true if both $f(x)$ and $g(x)$ are linear functions, particularly if 
$$f(x) = x.$$
Let us check that $f(x) = x$ is a solution:
$$f(xy + f(x)) = f(f(x)\cdot f(y)) + x \Leftrightarrow 
f(xy + f(x)) = xy + x \text{ and }f(f(x)f(y)) + x = xy + x \text{ (TRUE). }$$
A: We want to find all functions, continuous or not,
$\,f:\mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}\,$
such that for all $\,x\,$ and $\,y\,$ positive reals
$$ f(x y + f(x)) = f(f(x) f(y)) + x. \tag{1}$$
Now use equation $(1)$ with $\,y,x\,$ instead of
$\,x,y\,$ which gives
$$ f(x y + f(y)) = f(f(x) f(y)) + y. \tag{2}$$
Solving for $\,f(f(x)f(y))\,$ in both equations gives
$$ f(x y + f(y)) - y = f(x y + f(x)) - x.\tag{3}$$
Now suppose $\,f(x) = f(y).\,$ Equation $(3)$
implies that $\,x = y\,$ which proves $\,f\,$ is
one-to-one. 
Given $\,x>0,\,$ suppose $\,f(x)<x.\,$ 
Then we solve for $\,y>0\,$ in
$$ xy+f(x)=x. \tag{4} $$
Apply $\,f\,$ to both sides to get 
$$ f(x y + f(x)) = f(x). \tag{5}$$
Combine with equation $(1)$ to get
$$ f(x) = f(f(x) f(y)) + x. \tag{6}$$
This implies that $\,f(x) > x\,$
which contradicts our assumption $\,f(x) < x.\,$
Thus $\,f(x)\ge x\,$ for all $\,x>0.\,$
The obvious solution is $\,f(x)=x\,$ for all $\,x>0\,$
so now the question is how to prove $\,f(x)>x\,$ is impossible.
