parmenides51 wrote: Find all functions $ f: \mathbb {R} ^ + \to \mathbb {R} ^ + $ such that for any $ x, y \in \mathbb {R} ^ + $ the following equality holds: \[f (x) f (y) = f \left (\frac {xy} {x f (x) + y} \right). \]$ \mathbb {R} ^ + $ denotes the set of positive real numbers. Let $P(x,y)$ be the assertion $f(x)f(y)=f(\frac{xy}{xf(x)+y})$ 1) The only injective solutions are $f(x)=1+\frac ax$ Comparing $P(x,y)$ and $P(y,x)$, we get $f(\frac{xy}{xf(x)+y})=f(\frac{xy}{yf(y)+x})$ And so (injectivity) $xf(x)-x=yf(y)-y$ and so $xf(x)-x=a$ constant And so $\boxed{\text{S1 : }f(x)=1+\frac ax\quad\forall x>0}$ which indeed is a solution, whatever is $a>0$ Q.E.D. 2) The only non injective solution is $f\equiv 1$ 2.1) $f(x)\ge 1$ $\forall x>0$ If $f(x)<1$ for some $x>0$, then $P(x,x(1-f(x)))$ $\implies$ $f(x)=1$, contradiction. Q.E.D. 2.2) $f(x)\le f(y)$ $\forall x>y>0$ Let $x>y>0$ : $P(x,\frac{xyf(x)}{x-y})$ $\implies$ $f(x)f(\frac{xyf(x)}{x-y})=f(y)$ Q.E.D. 2.3) $f\equiv 1$ Since non injective, let $a>b>0$ such that $f(a)=f(b)=c$ $P(b,x)$ $\implies$ $cf(x)=f(\frac{bx}{bc+x})$ $P(a,\frac x{1+\frac {a-b}{abc}x})$ $\implies$ $cf(\frac x{1+\frac {a-b}{abc}x})=f(\frac{bx}{bc+x})$ And so $f(x)=f(\frac x{1+\frac {a-b}{abc}x})$ $\forall x>0$ And so (using 2.2 above), $f(x)$ constant over $\left[\frac x{1+\frac {a-b}{abc}x},x\right]$ Setting $x\to+\infty$, this means $f(x)$ constant over $\left(\frac{abc}{a-b},+\infty\right)$ Plugging this back in original equation with for example $x=y$ great enough, the constant is $1$ and so $f(x)=1$ $\forall x>A$ for some real $A>0$ But $f(x)=1$ implies, using $P(x,x)$ : $f(\frac x2)=1$ And so $\boxed{\text{S2 : }f(x)=1\quad\forall x>0}$ which indeed is a solutoin Q.E.D.