SELECTION EXAMINATION 2019

For $x=y=z=1$ we have $f(f(1))=1$, and hence for $x=y=z=f(1)$ we have $$3f(f(1))=3f(1{)}^2\Rightarrow f(1{)}^2=1\Rightarrow f(1)=1.$$ For $y=z=1$, we have for every $x\in {\mathbb{R}}_{+}$: $$\begin{array}{rl}f(f(x))+f(x)& =2x\\ f(f(x))& =2x-f(x)\end{array}(1)$$ For $z=1$ from the given relation we get: $$f(xf(y))+f(y)+f(f(x))=xy+y+x$$ And from (1) we find: $$f(xf(y))=xy+y-x+f(x)-f(y),(2)$$ for all $x,y\in {\mathbb{R}}_{+}$. Putting to (2) the $f(x)$ in the place of $x$ we have, from (1), for all $x,y\in {\mathbb{R}}_{+}$: $$f(f(x)f(y))=f(x)y+y+2x-2f(x)-f(y)(3)$$ and interchanging $x$ and $y$ we get the relation $$f(f(y)f(x))=f(y)x+x+2y-2f(y)-f(x)(4)$$ From (3) and (4) we have, for all $x,y\in {\mathbb{R}}_{+}$: $$(f(x)-1)(y-1)=(f(y)-1)(x-1),(5)$$ From which for $y=2$ we find $$f(x)=cx-c+1=c(x-1)+1(6)$$ where $c=f(2)-1$. By substitution to the given equation we have $$c^2(xy+yz+zx-x-y-z)+c(x+y+z)-3c+3=xy+yz+zx,$$ from which for $x=y=z$ we find $$\begin{gathered} 3(c^2-1)x^2+3c(1-c)x+3(1-c)=0 \\ \iff c^2-1=0,c(1-c)=0,1-c=0\iff c=1. \end{gathered}$$ $c=1$, and hence: $f(x)=x$, for every $x>0$.