The only solutions are $f(x)=1/x$ and $f(x)=1$, which can both easily be seen to work. Now, we prove that no other solutions exist. Let $S=\{x \in \mathbb{R} : f(x)=1 \}$. We have the following Claims. Claim 1: $1-1/x+f(x) \in S$ for all $x>1$. Proof: Indeed, we may take $x \rightarrow 1/(1-y)$ in the given equation to obtain $f(1-y+f(1/y))=1,$ and so $f(1-1/x+f(x))=1$ for all $x>1,$ as desired $\blacksquare$ Claim 1: If $u \in S$, then $f(ux)=f(x)$ for all $x$. Proof: Note that by Claim 1, $S$ is non-empty. Take an $u \in S$. Put $y=1/u$ in the given equation to obtain that $f(x/u+1)=f(x)f(1/x+1)$. Then, if we put $x \rightarrow u/x$ in this equation, $f(u/x)=\dfrac{f(1/x)+1}{f(x/u+1}=\dfrac{1}{f(x)}$ Hence, $f(u/x)=\dfrac{1}{f(x)}$ for all $x$. Hence, if we take $x=u$ here we obtain that $f(1)=1$. Hence, $f(1/x)=\dfrac{1}{f(x)},$ implying that $f(u/x)=f(1/x)$ for all $x$, i.e. $f(ux)=f(x)$ for all $x$, as desired $\blacksquare$ Claim 3: If $S \neq \{1 \}$, then $f$ is (eventually) periodic. Proof: Note that from Claim 2, if $u \in S$ then $f(xu)=f(x),$ hence $f(1/u)=1$. Thus, by putting $y=u$ in the initial equation we obtain $f(xu+1)=f(x)f(1/x+f(1/u))=f(x)f(1/x+1),$ and so $f(xu+1)=f(x)f(1/x+1)=f(x \cdot 1+1)=f(x+1),$ that is $f(xu+1)=f(x+1)$ for all $x$. Now, if we put $x \rightarrow xu$ in the initial equation, we obtain $f(\dfrac{1}{xu}+\dfrac{1}{f(y)})=\dfrac{f(xyu+1)}{f(x)}=\dfrac{f(xy+1)}{f(x)}=f(\dfrac{1}{x}+\dfrac{1}{f(y)}),$ Hence, if we put $x \rightarrow 1/x$ here, we obtain $f(\dfrac{x}{u}+\dfrac{1}{f(y)})=f(x+\dfrac{1}{f(y)}),$ and so by Claim 2, $f(x+\dfrac{1}{f(y})=f(x+\dfrac{u}{f(y)})$. Now, if $S \neq \{1 \},$ we may take a $u \neq 1,$ and so $f$ is (eventually) periodic, as desired $\blacksquare$ Back to the problem, we have two cases regarding set $S$. Case 1: $S \neq \{1 \}$. Then, $f$ is periodic with period $T>0$. By taking $y \rightarrow y+T$ in the given equation, we obtain $f(xy+1)=f(x(y+T)+1),$ and so by taking $x \rightarrow x/T$ and $y \rightarrow yT/x,$ we obtain $f(y+1)=f(y+x+1),$ hence by swapping $x,y$ we obtain that $f$ is constant in $>N$ inputs, for some $N>0$. However, since $f(x)=f(xu)$ and $u \neq 1,$ we obtain that $f(x)$ is constant for all $x$ (indeed, if WLOG $u>1$, then for all $x$ there exists a $k$ such that $xu^k>N$). Hence, $f \equiv 1,$ thus obtaining the second claimed solution. Case 2: $S=\{1 \}$. Then, by Claim 1 we immediately get that $f(x)=1/x$ for all $x>1$, and since $f(1/x)=1/f(x)$ this implies that $f(x)=1/x$ for all $x$, thus obtaining the first claimed solution.

The only solutions are $f(x) \equiv x^{-1}$ and $f(x)\equiv 1$. Let $P(x,y)$ denote the assertion in the problem. Then $P(x, \frac{x-1}{x})$ imply that $$ 1 = f( x^{-1} + f(\frac{x}{x-1}))$$ Case 1: $f$ injective. Then $f(\frac{x}{x-1})+x^{-1}=c$ implies that for all $y=\frac{x}{x-1} > 1$, $f(y) = c-1+\frac 1y$. It's easy to see that $c=1$, so $f(x)=\frac 1x$ for all $x\ge 1$. Now plug $x,y<1$ to see that $f(x) = x^{-1}$ for $x<1$ as well. Case 2: $f$ not injective. Say $f(a)=f(b)$ where $a<b$, then $P(x, 1/a), P(x,1/b)$ imply that $f(z+1)=f(z(b/a)+1)$ for all $z\in \mathbb{R}^+$ where $z = \frac{x}{b}$. Then $f(z+1)=f(cz+1)$ for any $c<b/a$ since $\frac{z(b/a)+1}{z+1}$ can take on any value in $(1, b/a)$, and setting $z$ such that $\frac{z(b/a)+1}{z+1}=c$ implies $f(z+1)\equiv f(cz+1)$, so we can see that $f(x) = k$ for some constant $k$ for all $x>1$. It's easy to prove that $k=1$ and $f(x)\equiv 1$ from here.

The solutions are $f(x)=\tfrac{1}{x}$ and $f \equiv 1$, which both clearly work. We now prove that no other solutions. Let the assertion be $P(x,y)$. By $P(2,\tfrac{1}{2})$, we find that there exists some $c$ such that $f(c)=1$. Suppose that $f$ is not injective, so there exists some $a \neq b$ such that $f(a)=f(b)$. Then by comparing $P(x,\tfrac{1}{a})$ and $P(x,\tfrac{1}{b})$, we have $f(\tfrac{x}{a}+1)=f(\tfrac{x}{b}+1)$ for all $x$. Thus define $S=\{r : f(x+1)=f(rx+1)~\forall x \in \mathbb{R}^+\}$, so $1 \in S$, $S$ is closed under multiplication and division, and $\tfrac{a}{b} \in S$. But then for all $r \in S$ and $x \in \mathbb{R}^+$, we also have $$\frac{rx+1}{x+1} \in S.$$As $x$ ranges across the positive reals, this fraction takes on every value between $1$ and $r$. Therefore by picking some arbitrarily large or small element of $S$ (which exists because $S$ is closed under multiplication and division), we find that every real is in $S$, i.e. there exists a constant $t$ such that for all $x>1$, $f(x)=t$. From $P(x,\tfrac{1}{c})$ where $x$ is arbitrary, we find that $f(>1)=f(x)f(>1) \implies f(x)=1$, yielding the solution $f \equiv 1$. On the other hand, if $f$ is injective, then from $P(c,\tfrac{1}{y})$ and injectivity we obtain $$\frac{c}{y}+1=\frac{1}{c}+f(y).$$Putting $y=c$ here, we then obtain $c=1$, so $f(y)=\tfrac{1}{y}$, which is our other solution, so we're done. $\blacksquare$

Quote: But then for all $r \in S$ and $x \in \mathbb{R}^+$, we also have $$\frac{rx+1}{x+1} \in S.$$ Can you prove this?

tadpoleloop wrote: Quote: But then for all $r \in S$ and $x \in \mathbb{R}^+$, we also have $$\frac{rx+1}{x+1} \in S.$$ Can you prove this? This is merely a consequence of $f(a)=f(b) \implies a/b \in S$. By the definition of $S$, for any $r \in S$ we have $f(rx+1)=f(x+1)$, and the conclusion follows

The only solutions are $\boxed{f\equiv 1}$ and $\boxed{f(x) = \frac{1}{x}}$. These clearly work. Now we prove they are the only solutions. Let $P(x,y)$ denote the given assertion. The only constant solution is $1$, so assume $f$ is nonconstant. By $P\left( x, \frac{1}{y} \right)$, we see \[ f\left( \frac{x}{y} + 1 \right) = f(x) f\left( \frac{1}{x} + f(y) \right) \]Denote this as $Q(x,y)$. Claim: If $f(a) = f(b)$, then $f(x + 1) = f\left( \frac{a}{b} \cdot x + 1 \right)$ for any $x>0$. Proof: Comparing $Q(x,a)$ and $Q(x,b)$ gives $f(1 + x/a ) = f(1 + x/b)$, so setting $x \to ax$ gives the desired result. $\square$ Claim: $f$ is injective. Proof: Suppose not. Fix two numbers with an equal output and ratio equal to $c>1$. We have $f(x+1) = f(cx + 1)$ for all $x$. Let $S$ be the set of $x$ with $f(x) = f(2)$. If $f(x) = f(y)$, then $Q(x,y)$ compared with $Q(x,x)$ gives that $\frac{x}{y} + 1\in S$. Hence $\frac{cx + 1}{x +1} + 1, \frac{x+1}{cx + 1} + 1 \in S$, which implies everything in the interval $\left[\frac{1}{c} +1, c+1\right]$ is in $S$ (the endpoints because the two numbers with ratio $c$ and equal output also have ratio $\frac{1}{c}$ and $2\in S$ is obvious). Notice that ratios between numbers in the interval can be anything from $\frac{1}{c} $ to $c$, so $f(x+1) = f(kx + 1)$ for any $\frac{1}{c} \le k\le c$. We see that $f$ is constant over the interval $[c + 1, c^2 + 1], [c^2 + 1, c^3 + 1], \ldots,$, so everything $[c+1, c^n + 1]$ for any positive integer $n$ is in $S$, which means that $y\in S\forall y \ge c + 1$. Now since $f(x) = f(cx + 1)$, we can repeatedly apply this to any positive $x$ and get that that $x\in S$, so $f$ is constant, contradiction. $\square$ Now we have for any $x>1$,\[ P\left(x, \frac{x-1}{x} \right) : f\left( \frac{1}{x} + f\left( \frac{x}{x-1}\right) \right) = 1,\]so $\frac{1}{x} + f\left( \frac{x}{x-1} \right) = r$ for some constant $r$. Now setting $x\to \frac{x}{x-1}$, we get that $f(x) = r - \frac{x-1}{x} = (r-1) + \frac{1}{x} $ for any $x>1$. If $r\ge 1$, then $1 = f(r) = (r - 1) + \frac{1}{r}$, so $r + \frac{1}{r} = 2\implies r = 1$. If $ r < 1$, then we have that $r > \frac{1}{x}$ for any $x > 1$, but choosing $1 < x < \frac{1}{r}$ gives a contradiction. Therefore, $r = 1$, so $f(x) = \frac{1}{x}$, as desired.

The solutions are $f\equiv 1$ and $f\equiv \dfrac{1}{x}$. $P\left(x,\frac{1}{y}\right) \implies f\left(\frac{x}{y} + 1 \right) = f(x) f\left(\frac{1}{x} + f(y)\right)$. Now let $f(a) = f(b)$, $a> b$. Comparing $P\left(x,\frac{1}{a}\right)$ and $P\left(x,\frac{1}{b}\right)$ gives $ f\left(\frac{x}{a} + 1\right) = f\left(\frac{x}{b}+1\right)\implies f(x+1) = f(rx + 1)$ where $r=\frac{a}{b} > 1$. Now fix $k\in \mathbb{R}^+$ arbitrarily such that $k \in (1,r)$ we can choose $x$ such that, $\dfrac{rx+1}{x+1} = k \iff x = \dfrac{k-1}{r-k}$. We then get that $f(x+1) = f(kx+1)$ for any $k\in (1,r)$. Note that we can also easily change $r$ to $r^2$ or some higher power which would help us in covering the entire $\mathbb{R}^+$. Now, \[ f(x+1) = f(kx+1) = f(kx^2 + 1) = \cdots \]which forces $f$ to be injective and otherwise constant $=1$ over $(1,\infty)$ and thus over $(0,\infty)$. Now for $y\in (0,1)$. $P\left(\frac{1}{1-y},y\right) \implies f\left(\dfrac{1}{\frac{1}{1-y}} + f\left(\frac{1}{y}\right)\right) = 1$. So $f(y) = \frac{1}{y} + c$ for all $y \in (1,\infty)$. Now $P\left(2,\frac{1}{2}\right) \implies c=0$. So, $f(y) = \frac{1}{y} \; \forall \; y\in (1,\infty)$. Now for some $x\in (0,1)$, \[ P(x,x) \implies \dfrac{1}{x^2 + 1} = f(x) f\left(\frac{1}{x} + x\right) \implies \dfrac{1}{x^2 + 1} = f(x) \cdot \dfrac{x}{x^2 + 1}\implies f(x) = \frac{1}{x}. \] Now choose $y\neq 1$. Then $P(1,y) \implies f(1) = 1$, and so, we conclude that $f(x) = \frac{1}{x} \; \forall \; x\in \mathbb{R}^+$.

The answer is $f(x) = \boxed{\tfrac{1}{x}, 1}$, which work. Let the given assertion be denoted as $P(x,y)$. It is clear that $1$ is the only constant solution, so assume $f$ is nonconstant. By $P(2,\tfrac{1}{2})$, there exists a value $c$ such that $f(c)=1$. Claim: $f$ is injective. Proof: Suppose not. This means there exists two distinct numbers $a$ and $b$ with ratio $r>1$, such that $f(a)=f(b)$. Comparing $P(x, \tfrac{1}{a})$ and $P(x, \tfrac{1}{b})$ gives \[f\left(\frac{x}{a}+1 \right) = f\left(\frac{x}{b}+1 \right),\] at which point we can map $x \to ax$ so that \[f(x+1) = f(rx+1).\] Notice that if we take $k$ such that $\tfrac{rx+1}{x+1}=k$, we get $f(x+1) = f(kx+1)$. But, $k$ ranges throughout $(1,r)$, and we can tweak the factor of our initial mapping in order to cover the intervals $(r^n, r^{n+1})$ for $n \in \mathbb{Z}$. This implies $f$ is constant, a contradiction. $\square$ Now that we know $f$ is injective, utilize $c$ by plugging in $P(c,\tfrac{1}{c})$ to get \[f(2) = f(c)f\left(\frac{1}{c}+f(c) \right)\]\[\implies 2 = 1+\frac{1}{c}.\] This means that $c=1$. Finally, $P(c,\tfrac{1}{y})$ yields \[\frac{c}{y}+1 = \frac{1}{c} + f(y), \] which gives us our other solution.

Let $P(x,y)$ denote the functional equation. Let us define the set $S=\{x:f(x)=1\}$. First we'll show $S$ is non-empty. Indeed there exists positive $x$ such that $$1+xy=\frac 1x +f\left(\frac 1y\right) \iff x^2y+x\left (1-f\left(\frac 1y\right)\right)-1=0,$$by, say, Descartes' Rule of Signs. So, the functional equation implies that $f(1/x)=1$, as desired. Let $t\in S$. Then \begin{align*} P(x,1/t) & \ \Rightarrow \ f\left (1+\frac xt\right)=f(x)f\left(\frac 1x+1\right) \\ P(t/x,1/x) & \ \Rightarrow \ f\left (1+\frac 1x\right)=f\left (\frac tx\right)f\left(\frac xt+1\right) \qquad (1) \end{align*}So, $$f\left (1+\frac 1x\right)=f\left (\frac tx\right)f\left(\frac xt+1\right)=f\left(\frac tx\right)f(x)f\left(\frac 1x+1\right) \Rightarrow f\left(\frac tx\right)f(x)=1 \qquad (2)$$If we put $x=1$ in $(2)$ then we get $f(1)=1$, so $1\in S$. Since all the work we did above holds for any arbitrary element of $S$, it follows that $$f\left(\frac 1x\right)=\frac{1}{f(x)}, \quad f\left(\frac 1x+1\right)=\frac{f(1+x)}{f(x)} \qquad (3)$$Notice that $(2)$ now implies $$f\left(\frac tx\right)=f\left(\frac 1x\right) \Rightarrow f(tx)=f(x)$$for all $x>0$ and $t\in S$. In particular, it is now clear that $1/t\in S$ if $t\in S$. $(1)$ gives us $$\frac{f(1+x)}{f(x)}=f\left (\frac 1x\right)f\left(\frac xt+1\right)\Rightarrow f(1+x)=f\left(1+\frac xt\right) \Rightarrow f(1+xt)=f(1+x)$$ Assume that there is a $t\in S$ such that $t\neq 1$. WLOG we can assume $t<1$. Since $t\in S$, $t^n\in S$ for any integer $n$. So we have $$f(xt^n+1)=f(x+1)=f(xt^n+t^n)\Rightarrow f(x+1)=f(x+t^n).$$So, $f(y)=f(y+1-t^n)$ for all $y>t^n$. $P(x,y+1-t^n)$ and $P(x,y)$ implies that $$f(1+x(y+1-t^n))=f(1+xy) \Rightarrow f(2)=f\left(2+\frac{1-t^n}{y}\right),$$so it follows that $f(x+2)=f(2)$ for all $x<t^{-n}-1$. But as $n\to +\infty$, $t^{-n}-1\to +\infty$, so $f(x+2)=f(2)$ must hold for all $x$. Next, $$P(1/2,2y)\implies f(y+1)=\frac{1}{f(2)}f(2)=1.$$So $f(x+1)=1$ for all $x$ but then $(3)$ implies that $f(x)=1$ for all $x$. Clearly this function works. It remains to consider the case $S=\{1\}$. We will show that $f$ is injective. Indeed, suppose $f(a)=f(b)$. Then by $P(x,a)$ and $P(x,b)$ we see that $f(1+ax)=f(1+bx)$. So, $$f\left (1+\frac ab\right)=f(2)=f\left (1+\frac ba\right),$$and from this $(3)$ gives us $f(b/a)=1$. Thus, $b/a \in S$, so $a=b$, as required. So by injectivity $$P(1,y)\Rightarrow f(1+y)=f(1+1/f(y))\Rightarrow 1+y=1+1/f(y)\Rightarrow f(y)=1/y \quad \forall y$$Conversely, $f(x)=1/x$ for all $x$ clearly works. So we're done.

kamatadu wrote: The solutions are $f\equiv 1$ and $f\equiv \dfrac{1}{x}$. $P\left(x,\frac{1}{y}\right) \implies f\left(\frac{x}{y} + 1 \right) = f(x) f\left(\frac{1}{x} + f(y)\right)$. Now let $f(a) = f(b)$, $a> b$. Comparing $P\left(x,\frac{1}{a}\right)$ and $P\left(x,\frac{1}{b}\right)$ gives $ f\left(\frac{x}{a} + 1\right) = f\left(\frac{x}{b}+1\right)\implies f(x+1) = f(rx + 1)$ where $r=\frac{a}{b} > 1$. Now fix $k\in \mathbb{R}^+$ arbitrarily such that $k \in (1,r)$ we can choose $x$ such that, $\dfrac{rx+1}{x+1} = k \iff x = \dfrac{k-1}{r-k}$. We then get that $f(x+1) = f(kx+1)$ for any $k\in (1,r)$. Note that we can also easily change $r$ to $r^2$ or some higher power which would help us in covering the entire $\mathbb{R}^+$. Now, \[ f(x+1) = f(kx+1) = f(kx^2 + 1) = \cdots \]which forces $f$ to be injective and otherwise constant $=1$ over $(1,\infty)$ and thus over $(0,\infty)$. Now for $y\in (0,1)$. $P\left(\frac{1}{1-y},y\right) \implies f\left(\dfrac{1}{\frac{1}{1-y}} + f\left(\frac{1}{y}\right)\right) = 1$. So $f(y) = \frac{1}{y} + c$ for all $y \in (1,\infty)$. Now $P\left(2,\frac{1}{2}\right) \implies c=0$. So, $f(y) = \frac{1}{y} \; \forall \; y\in (1,\infty)$. Now for some $x\in (0,1)$, \[ P(x,x) \implies \dfrac{1}{x^2 + 1} = f(x) f\left(\frac{1}{x} + x\right) \implies \dfrac{1}{x^2 + 1} = f(x) \cdot \dfrac{x}{x^2 + 1}\implies f(x) = \frac{1}{x}. \] Now choose $y\neq 1$. Then $P(1,y) \implies f(1) = 1$, and so, we conclude that $f(x) = \frac{1}{x} \; \forall \; x\in \mathbb{R}^+$. I didn’t understand this part: Now fix $k\in \mathbb{R}^+$ arbitrarily such that $k \in (1,r)$ we can choose $x$ such that, $\dfrac{rx+1}{x+1} = k \iff x = \dfrac{k-1}{r-k}$. We then get that $f(x+1) = f(kx+1)$ for any $k\in (1,r)$. Note that we can also easily change $r$ to $r^2$ or some higher power which would help us in covering the entire $\mathbb{R}^+$. Now, \[ f(x+1) = f(kx+1) = f(kx^2 + 1) = \cdots \]which forces $f$ to be injective and otherwise constant $=1$ over $(1,\infty)$ and thus over $(0,\infty)$. Can you explain it?