Tintarn wrote: Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+\to\mathbb{R}^+$ such that \[ \frac{f(a)}{1+a+ca}+\frac{f(b)}{1+b+ab}+\frac{f(c)}{1+c+bc} = 1 \]for all $a,b,c \in \mathbb{R}^+$ that satisfy $abc=1$. Replacing $(a,b,c)$ by $(x,y,\frac 1{xy})$, property becomes Assertion $P(x,y)$ : $yf(x)+f(y)+xyf(\frac 1{xy})=xy+y+1$ $\forall x,y>0$ Subtracting $P(y,x)$ from $P(x,y)$, we get $yf(x)+f(y)-xf(y)-f(x)=y-x$ Setting there $y=2$, we get $f(x)=x(f(2)-1)+(2-f(2))$ Plugging $f(x)=ax+b$ in $P(x,y)$, we get $a+b=1$ and so $\boxed{f(x)=ax+1-a\quad\forall x>0}$, which indeed fits, whatever is $a\in[0,1]$ (this condition to match $f(x)>0$ $\forall x>0$)

First get rid of $abc=1$ by taking $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ so that the F.E. becomes: \[y \cdot f \left(\frac{x}{y} \right)+z \cdot f \left(\frac{y}{z} \right)+x \cdot f \left(\frac{z}{x} \right)=x+y+z \]Denote $P(x,y,z)$ as the assertion of this F.E., then $P(1,1,1)$ gives $f(1)=1$ clearly, $P(x,t,t)$ gives: \[t \cdot f \left(\frac{x}{t} \right)+x \cdot f \left(\frac{t}{x} \right)=x+t \]The way we use this on the original F.E. to get that (call it $Q(x,y,z)$): \[ z \cdot f \left(\frac{y}{z} \right)+x \cdot f \left( \frac{z}{x} \right)=x \cdot f \left(\frac{y}{x} \right)+z \]Take $(y,z) \to (xy,xz)$ to get that $z \cdot f \left(\frac{y}{z} \right)+f(z)=f(y)+z$, now take $y=xz$ here again to get that $zf(x)+f(z)=f(xz)+z$ for all positive reals $x,z$ which now by symetry gives $zf(x)+f(z)+x=xf(z)+f(x)+z$ and clearly from here setting $z$ constant gives $f(x)=ax+b$ for some $a,b>0$, and replacing gives $b=1-a$ and $a \le 1$, thus we are done .