GeorgeRP wrote: Find all functions \(f:\mathbb{R}^{+} \to \mathbb{R}^{+}\) such that \[ f(x) > x \ \ \text{and} \ \ f(x-y+xy+f(y)) = f(x+y) + xf(y) \]for arbitrary positive real numbers \(x\) and \(y\). Let $P(x,y)$ be the assertion $f(x-y+xy+f(y))=f(x+y)+xf(y)$ Let $x>2$ : $P(2,x-2)$ $\implies$ $f(x+f(x-2))=f(x)+2f(x-2)$ Which is $f(u)-2u=f(x)-2x$ where $u=x+f(x-2)>x$ and then : $P(u,x)$ $\implies$ $f(u+x+ux+(f(x)-2x))=f(x+u)+uf(x)$ $P(x,u)$ $\implies$ $f(u+x+ux+(f(u)-2u))=f(x+u)+xf(u)$ $=f(x+u)+x(f(x)-2x+2u)$ And so, subtracting, $uf(x)=x(f(x)-2x+2u)$, which is $(u-x)(f(x)-2x)=0$ And so $f(x)=2x$ $\forall x>2$ Then, for any $x>0$, $P(3,x)$ becomes $2(3-x+3x+f(x))=2(x+3)+3f(x)$ And so $\boxed{f(x)=2x\quad\forall x>0}$, which indeed fits

A shifted version of USEMO 2020/4.

Now, if we prove that f(y) can‘t be bigger than, we will be ready, because Let‘s call this equation P(x,y) P(2.y-f(y)/y; y) gives us f(2.y-f(y)/y -y+2.y-f(y)+f(y))=f(2.y-f(y)/y +y)+2.y-f(y)/(y) .f(y) The left side and the first expression are same, then we have that 2.y-f(y)/(y) .f(y)=0 The we have that either f(y)=0, or 2.y=f(y) The one problem of this is that we aren‘t sure that 2.y>=f(y) I think I have a solution for this(by contradiction) Here it is, assume that f(y)>2.y P(1,y) give us f(1-y+y+f(y))=f(y+1)+f(y) We have that: f(f(y)+1)=f(y+1)+f(y) The left side is bigger than 2.f(y)+2 2.f(y)+2<=f(y+1)+f(y) Therefore: f(y)+2<=f(y+1) After we put this: y->infinite, we get maybe contradiction with 2<=0, I don’t think that is clearly true, but…. Now, we are ready with 2.y>=f(y), because above