a_507_bc wrote: Find all functions $f:\mathbb{R^{+}} \rightarrow \mathbb{R^{+}}$, such that $$xf(1+xf(y))=f(f(x)+f(y))$$for all positive reals $x, y$. Let $P(x,y)$ be the assertion $xf(1+xf(y))=f(f(x)+f(y))$ $P(\frac x{f(x+1)},x+1)$ $\implies$ $x=f(f(\frac x{f(x+1)})+f(x+1))$ and so $f(x)$ is surjective and we can rewrite $P(x,y)$ as New assertion $Q(x,y)$ : $xf(1+xy)=f(f(x)+y)$ $Q(1,x)$ $\implies$ $f(1+x)=f(f(1)+x)$ and so $f(x)=f(x+T)$ (where $T=|f(1)-1|$) for all $x$ great enough Then, comparaison of $Q(x,1)$ with $Q(x+T,1)$ implies $x=x+T$ for all $ x$ great enough and so $T=0$ and $f(1)=1$ Let $x\ne 1$. If $\frac{f(x)-1}{x-1}>0$, then $Q(x,\frac{f(x)-1}{x-1})$ $\implies$ $x=1$, impossible. So $\frac{f(x)-1}{x-1}\le 0$ $\forall x\ne 1$ If $f(u)=1$ for some $u>1$, then $Q(u,\frac {u-1}u)$ $\implies$ $u=f(1+\frac {u-1}u)$, in contradiction with previous line. So $f(x)<1$ $\forall x>1$ If $f(f(x))<x$, then $Q(f(x),x-f(f(x)))$ implies $f(1+f(x)(x-f(f(x))))=1$, in contradiction with previous result. So $f(f(x))\ge x$ $\forall x>0$ With $f(1+xy)<1$, $Q(x,y)$ $\implies$ $f(f(x)+y)<x$ and so $f(f(x)+y)<f(f(x))$ and surjectivity implies $f(x+y)<f(x)$ and $f(x)$ is stricly decreasing (and so injective). $f(f(x))\ge x$ implies (moving $x\to f(x)$) $f(f(f(x)))\ge f(x)$ If $f(f(x))>x$ for some $ x$, "stricly decreasing" implies $f(f(f(x)))< f(x)$, contradiction with above. So $f(f(x))=x$ $\forall x>0$ Let $x,y<1$ (so that $f(x)>1$ and $f(y)>1$ $Q(x,\frac{f(y)-1}x)$ $\implies$ $xy=f(f(x)+\frac{f(y)-1}x)$ Switching there $x,y$ and using injectivity, we get $f(x)+\frac{f(y)-1}x=f(y)+\frac{f(x)-1}y$ and so $f(x)=\frac ax+b$ $\forall x<1$ for some $a>0,b$ If $x>1$, we have $f(x)<1$ and so $x=f(f(x))=\frac a{f(x)}+b$ and so $f(x)=\frac a{x-b}$ $\forall x>1$ $Q(x,1)$ $\implies$ $xf(1+x)=f(1+f(x))$ and so $\frac {ax}{1+x-b}=\frac a{1+f(x)-b}$ and so $f(x)=\frac{1-b}x+b$ $\forall x$ If $x<1$, since we already got $f(x)=\frac ax+b$, we conclude $a=1-b$ If $x>1$, since we already got $f(x)=\frac a{x-b}$, we conclude $\frac {1-b}{x-b}=\frac{1-b}x+b$ $\forall x>1$ and so $b=0$ And solution $\boxed{f(x)=\frac 1x\quad\forall x>0}$, which indeed fits

a_507_bc wrote: Find all functions $f:\mathbb{R^{+}} \rightarrow \mathbb{R^{+}}$, such that $$xf(1+xf(y))=f(f(x)+f(y))$$for all positive reals $x, y$. By setting $x=\frac{t}{f(t+1)},y=t+1$, we get that $f$ is surjective. So we get the new functional equation \[ xf(1+xy)=f(f(x)+y). \]From here we will work with the new FE. If $f(x)>1$ for $x>1$ or $f(x)<1$ for $x<1$ setting $y=\frac{f(x)-1}{x-1}$ leads to the contradiction $x=1$. So $f(x)\leq1$ for $x>1$ and $f(x)\geq1$ for $x<1$. Setting $y=1-\frac{1}{x}$ for $x>1$ gives \[ xf(x)=f\left(\frac{1}{x}(xf(x)-1)+1\right). \]In this equation the assumtions $xf(x)>1$ or $xf(x)<1$ both lead to contradictions. So $f(x)=\frac{1}{x}$ for $x>1$. Now, choosing $y=1$ gives \begin{align*} x\cdot\frac{1}{1+x}=\frac{1}{f(x)+1}\\ f(x)=\frac{1}{x}. \end{align*}It is easy to check that $f(x)=\frac{1}{x}$ is a solution.