We have $f$ - nonconstant. a) if $y=1$, then $(y-1)(f(y)-1) = 0$. Suppose that $y\neq 1$. If $(y-1)(f(y)-1)>0$, then at $P(\frac{f(y)-1}{y-1},y)$ $ \rightarrow$ $y=1$ , contradiction. Hence, $(f(y)-1)(y-1)\le 0$. b) since a), if $y>1$, then $f(y)\le 1$ and if $y<1$ then $f(y)\ge1$. Suppose that $y>1$ . $P(\frac{y-1}{y}, y)$ $ \Rightarrow $ $ yf(y)=f(1-\frac{1}{y}+f(y))$ and $f(y)=\frac{1}{y}$. Hence \[ f(x+f(y))=yf(xy+1)=\frac{y}{xy+1} .\] So, $f(1)=1$ and \[ f(x)=\frac{1}{x} \] for all $x \in R^+ .$

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We have $f$ - nonconstant. a) if $y=1$, then the equality holds. Suppose that $ y\not= 1 $. If $ (y-1)(f(y)-1)>0 $, then at $ P(\frac{f(y)-1}{y-1},y) $ $ \rightarrow $ $ y=1 $ , contradiction. Hence, $ (f(y)-1)(y-1)\le 0 $. b) since a), if $y>1$, then $f(y)\le 1$ and if $y<1$ then $f(y)\ge1$. Suppose that $y>1$ . $P(\frac{y-1}{y}, y)$ $ \Rightarrow $ $ yf(y)=f(1-\frac{1}{y}+f(y))$ and $f(y)=\frac{1}{y}$. Hence \[ f(x+f(y))=yf(xy+1)=\frac{y}{xy+1} .\] So, $f(1)=1$ and \[ f(x)=\frac{1}{x} \] for all $x \in R^+ .$

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=533460

mathuz wrote: We have $f$ - nonconstant. a) if $y=1$, then $(y-1)(f(y)-1) = 0$. Suppose that $y\neq 1$. If $(y-1)(f(y)-1)>0$, then at $P(\frac{f(y)-1}{y-1},y)$ $ \rightarrow$ $y=1$ , contradiction. Hence, $(f(y)-1)(y-1)\le 0$. b) since a), if $y>1$, then $f(y)\le 1$ and if $y<1$ then $f(y)\ge1$. Suppose that $y>1$ . $P(\frac{y-1}{y}, y)$ $ \Rightarrow $ $ yf(y)=f(1-\frac{1}{y}+f(y))$ and $f(y)=\frac{1}{y}$. Hence \[ f(x+f(y))=yf(xy+1)=\frac{y}{xy+1} .\]So, $f(1)=1$ and \[ f(x)=\frac{1}{x} \]for all $x \in R^+ .$ Great solution

$P(\frac{y-1}{y}, y)$ $ \Rightarrow $ $ yf(y)=f(1-\frac{1}{y}+f(y))$ and $f(y)=\frac{1}{y}$ How?

@above if $f(y)<\frac{1}{y}$ then $1-\frac{1}{y}+f(y)<1\rightarrow yf(y)=f(1-\frac{1}{y}+f(y))\geq 1\rightarrow f(y)\geq \frac{1}{y}$, contradiction. If $f(y)>\frac{1}{y}$ then $1-\frac{1}{y}+f(y)>1\rightarrow yf(y)=f(1-\frac{1}{y}+f(y))\leq 1\rightarrow f(y)\leq \frac{1}{y}$, contradiction. So $f(y)=\frac{1}{y}$.

In mathuz's solution, why does $ P(\frac{f(y)-1}{y-1},y) \rightarrow y=1$ ?

Plops wrote: In mathuz's solution, why does $ P(\frac{f(y)-1}{y-1},y) \rightarrow y=1$ ? just substitute and because of the condition problem there is a contradiction

Olemissmath wrote: Let $f:\mathbb R^+ \to \mathbb R^+$ be a function such that: \[ x,y > 0 \qquad f(x+f(y)) = yf(xy+1). \, (*) \] Another approach a) Show that $(y-1)*(f(y)-1) \le 0$ for $y>0$. $$Ez$$ b) Find all such functions that require the given condition. Claim 1: $f$ is injective Proof: Assume $\exists \, a>b: f(a)=f(b)=c$ $- \, \frac{P(x;a)}{P(x;b)}: \frac{af(ax+1)}{bf(bx+1)}=1 \leftrightarrow \frac{f(ax+1)}{f(bx+1)}=\frac{b}{a}\leftrightarrow \frac{f(\frac{a}{b}.x+1)}{f(x+1)}=\frac{b}{a}$ - If $\exists \, \frac{b}{a}<y<1: \, f(y)>1, P\left (\frac{b\left [ f(y)-1 \right ]}{ay-b};y \right ): f\left ( \frac{b}{a}.something+1 \right )=yf(something+1) \rightarrow y=\frac{a}{b}$, a contradiction -Similarly, it's a contradiction if $\exists \, 1<y<\frac{a}{b}: \, f(y)<1$ So $f(y)=1, \, \forall y \in \left ( \frac{b}{a};1 \right ) \cap \left ( 1;\frac{a}{b} \right )$ Choose $x,y>0: \, \left\{\begin{matrix} y \in \left ( \frac{b}{a};1 \right ) \cap \left ( 1;\frac{a}{b} \right ) & & \\x+1 \in \left ( 1;\frac{a}{b} \right ) & & \\xy +1 \in \left ( 1;\frac{a}{b} \right ) & & \\ \end{matrix}\right.$ in $(*)$ then $y=1$, a contradiction Q.E.D $P(1;1):$ $f(1)=1$ Claim 2: $f$ is not increasing Proof: $\frac{P(xz;y)}{P(xy;z)}:\frac{f(xz+f(y))}{y}=\frac{f(xy+f(z))}{z}, (1)$ If $\exists y<z: f(y)<f(z)$ then let $x=\frac{f(z)-f(y)}{z-y}$ in $(1)$, we get $y=z$, a contradiciton Q.E.D Claim 3: $f$ is surjective Proof: Let $x= \frac{1}{y}, \, z=1$ in $(1)$, we get: $f\left ( \frac{1}{y}+f(y) \right )=y.f(1+f(1))$. Q.E.D From Claim 1, Claim 2 and Claim 3, we get: $f$ is continuous Let $x \rightarrow 0$ in $(*)$, we get: $f(f(y))=y$ From $(*)$, we get: $\frac{f(x+f(y))}{y}=\frac{f(y+f(x))}{x}\, \left ( =f(xy+1) \right ), \, (2)$ Let $x \rightarrow f(x)$ in $(2)$, we get: $\frac{f(x)}{y}=\frac{f(x+y)}{f(f(x)+f(y))} \Rightarrow \frac{f(x)}{y}=\frac{f(y)}{x}\leftrightarrow xf(x)=yf(y), \forall x,y>0$ $\rightarrow$ So $f(x)=\frac{a}{x} \, \forall x>0$ for some constant $a>0$ But $f(1)=1$ so $a=1$ So the only solution is: $$ f(x)=\frac{1}{x} \, \forall x>0$$