Clearly $f$ is surjective on $(1,\infty)$ and injective as if $f(z_1) = f(z_2)$, then $y=z_1-x$ and $y=z_2-x$ for any $x<z_1,z_2$ yield $f(xf(z_1)) = (z_1-x)f(x) + 1$, $f(xf(z_2)) = (z_2-x)f(x) + 1$ and hence $z_1=z_2$. To use these properly, take any $z>1$, consider $x_0$ with $z=f(x_0)$ and aim to make the right-hand side of the initial equation with an $f$ in order to use injectivity. To be precise, setting $x=x_0$ and $y=\frac{z-1}{z}$ yields $f(x_0f(x_0+\frac{z-1}{z})) = f(x_0)$, thus $f(x_0 + \frac{z-1}{z}) = 1$ and so $x_0 + \frac{z-1}{z} = k$ for some constant $k$, thus $$\mbox{If } x \mbox{ is such that }f(x) > 1, \mbox{ then } f(x) = \frac{1}{x+1-k}.$$Now, in the initial equation we have $f(xf(x+y)) = yf(x) + 1 > 1$, hence this equation becomes (with $f(k) = 1$) $$ \frac{1}{xf(x+y) + 1-k} = yf(x) + 1 $$Setting $y=k-x$ yields $1 = (x+1-k)((k-x)f(x) + 1) \Leftrightarrow f(x) = \frac{1}{x+1-k}$ for all $x<k$. Setting $x=k$ yields $\frac{1}{kf(k+y) + 1 -k} = y+1$ and hence $kf(z) + 1 - k = \frac{1}{z-k+1} \Leftrightarrow f(z) = \frac{(k-1)z-k^2+2k}{zk - k^2+k}$ for all $z>k$. Now from $\frac{1}{x+1-k} = f(x) > 0$ we have $x > k-1$ for all $0<x<k$, thus $k\leq 1$; and in $\frac{(k-1)z-k^2+2k}{zk - k^2+k} = f(z) > 0$ for large $z$ and fixed $k<1$ the numerator will be negative and the denominator will be positive, contradiction! Therefore $k=1$ and substituting above yields $f(x) = \frac{1}{x}$ for all of $x<1$, $x=1$ and $x>1$, i.e. $f(x) = \frac{1}{x}$ for all $x$ (which satisfies the given equation).

Very nice we prove that the only function which works is $f(x)=\dfrac 1x$ and it is easy to verify that this indeed works. Let $P(x,y)$ denote the given assertion. Firstly, fixing $x$ gives that $f$ is surjective over $(1, \infty)$. If $f(a)=f(b)$, then pick $c < \min\{a,b\}$, equating $P(c,a-c)$ and $P(c,b-c)$ gives $a=b$ hence $f$ is injective. Claim: $$f\left(x + \dfrac{1}{f(x)}\right) = \dfrac cx$$for some constant $c$ also from here onwards let this assertion denote $Q(x)$. Proof: Equating $P(1,1/f(1))$ and $P(x, 1/f(x))$ we have $$f(xf(x + 1/f(x)))=2=f(f(1+1/f(1)))$$and because of injectivity, $xf(x+1/f(x)) = f(1 + 1/f(1)) = c$. Claim: $f(1)=1$. Proof: $Q(c) \implies$ there exists $t$ such that $f(t)=1$. If $t > 1$ then $P(1,t-1) \implies f(t-1) = f(t-1) + 1$ which is a contradiction, hence $t \le 1$. Note that $Q(x)$ means that $f$ is completely surjective, if $t <1$ then choose $x_0$ such that $f(x_0)=\dfrac{1}{1-t}$ then $P(x_0,t) \implies f(x_0f(x_0 + t)) = tf(x_0)+1 = \dfrac{1}{1-t} = f(x_0)$ and because of injectivity it means $f(x_0 + t) = 1 = f(t)$ which is a contradiction. To conclude, $f(1)=1$. Note that $P(x,1-x) \implies f(x) = \dfrac 1x$ for all $x < 1$. From $f(1)=1$, $Q(1)\implies c=f(2)$, $P(1,1) \implies f(c) = f(f(2))=2$, $P(1/2 , 1/2) \implies f(1/2) = 2 = f(c) \implies c=1/2$. $P(1,x) \implies f(f(1+x))=1+x \implies f(f(x)) = x$ for all $x\ge 1$. Claim: $f(x) = \dfrac 1x$ for all $x \ge 1$. Proof: Choose $x\ge 1$, $$f(Q(x)) \implies x + \dfrac{1}{f(x)} = f(f(x + 1/f(x))) = f\left(\dfrac{1}{2x}\right)$$and since $x\ge 1 \implies \dfrac{1}{2x} <1$ thus $$x + \dfrac{1}{f(x)} = f\left(\dfrac{1}{2x}\right) = 2x \implies f(x) = \dfrac 1x$$and this solves the problem.

The answer is $f(x) \equiv \frac{1}{x}$. It is easy to see that it works. Let $P(x,y)$ be the given assertion. Claim 1: $f$ is injective. Proof: Suppose $f(a) = f(b)$. Fix a small $x$ (we only require $x < a,b$). Comparing $P(x,a-x)$ and $P(x,b-x)$ gives $$ (x-a)f(x) = (x-b)f(x) $$As $f(x) \ne 0$, hence $a=b$, as desired. $\square$ Call two pairs $(x_1,y_1)$ and $(x_2,y_2)$ similar (and denote $(x_1,y_1) \sim (x_2,y_2)$) if $$ y_1f(x_1) = y_2f(x_2) \qquad \qquad (1)$$Comparing $P(x_1,y_1)$ and $P(x_2,y_2)$ (and invoking $f$ is injective) gives $$ x_1f(x_1 + y_1) = x_2 f(x_2 + y_2)\qquad \qquad (2) $$whenever $(x_1,y_1) \sim (x_2,y_2)$. Basically, if $(1)$ is true $\iff$ $(2)$ is true. Note in some sense $(1)$ is just fixing the ratio of $y_1,y_2$. Claim 2: $f$ is strictly decreasing. Proof: Recall $f$ was injective. Assume on the contrary that $$x_1 < x_2 ~~ \text{and} ~~f(x_1) < f(x_2)$$Consider all $y_1,y_2$ such that $(1)$ is true. Then we have $$ x_1f(x_1 + y_1) = x_2 f(x_2 + y_2) $$But due to our assumptions, $y_1,y_2$ can be chosen such that $$ x_1 + y_2 = x_2 + y_2 $$So that forces $$ x_1 = x_2 $$which is a contradiction. $\square$ Claim 3: $f$ is continuous. Proof: (We basically vary $y$ in $P(x,y)$ to prove this) Fix $x$ and consider any $a$. It suffices to show (as we could take $x$ small) that $$ f(x+a) = \lim_{z \to a} f(x+z) $$Let $I$ be any of the below two sets (or intervals) $$\{z : 0 < a < x+a \} ~~,~~ \{z : a < z\}$$Pick any small $\epsilon > 0$. As $f$ is strict monotone, so it suffices to show set $$ f(x + I) - f(x+a) $$has an element $\in (-\epsilon,\epsilon)$. Assume on the contrary that $$|f(x+z) - f(x+a)| \ge \epsilon ~~ \forall ~ z \in I$$So there is a $\alpha > 0$ such that $$ |xf(x+a) - xf(x+z)| \ge \alpha ~~ \forall ~ z \in I $$As $f$ is strict monotone, so $\exists ~ \beta >0$ such that $$ \left\vert f(xf(x+a)) - f(xf(x+z)) \right\vert \ge \beta ~~ \forall ~ z \in I $$But then invoking $P(x,a)$ and $P(x,z)$ gives $$ \left\vert af(x) - zf(x) \right\vert \ge \beta ~~ \forall ~ z \in I $$But tending $z \to a$, the LHS tends to $0$, so we obtain our desired contradiction. $\square$ Look at $(1),(2)$ now. Fix $x_1,x_2$. Pick $y_1,y_2$ satisfying $(1)$ and tend $y_1,y_2 \to 0$. Using continuity of $f$ we obtain $$ x_1f(x_1) = x_2f(x_2) ~~ \forall ~ x_1,x_2$$It follows there exists a $c$ such that $$ f(x) \equiv \frac{c}{x} $$Now there are several ways to conclude $c=1$: Just plug in $f$ in $P(x,y)$. Note $f$ is always a involution. $P(1,y)$ gives $$yf(1) + 1 = f(f(1+y)) = 1+y$$Which forces $f(1) = 1$, hence $c=1$. This completes the proof. $\blacksquare$

We claim that $f\equiv \text{Id}^{-1}$, which clearly works. Denote the assertion by $P(x,y).$ Claim: $f$ is bijective. Proof. For injectivity, take $f(u)=f(v)$ and a small $w$ then by combining $P(w,u-w)$ and $P(w,v-w)$ we get $u=v.$ Then take $P(x,\tfrac{1}{f(x)})$ so that we get $xf(x+\tfrac{1}{x})$ constant. This is enough for surjectivity. $\blacksquare$ Pick an $a$ for $f(a)=1.$ If $a>1$ then $P(a-1,1)$ gives contradiction. If $a<1$ then take $b$ for $f(\tfrac{1}{b})=1-a.$ By $P(\tfrac{1}{b},a)$ we get contradiction. Note that $f(1)=1.$ For all $x>1$ by $P(1,x-1)$ and for all $x<1$ by $P(x,1-x)$ yields $f(x)=\tfrac{1}{x},$ as desired.

I hope it works Let $P(x,y)$ be the assertion of the given functional equation. It is not so hard to prove $f$ is bijective. Without a problem, let $y \rightarrow 0$. We have: $$f(xf(x))=1$$Obviously, $f$ is not constant. Hence we can conclude that $xf(x)$ is constant. Therefore, $f(x)=\frac{k}{x}$ for some positive real number $k$. Putting to the our equation, we get: $$\frac{x+y}{x}= \frac{ky+x}{x}$$ Hence we conclude $k=1$ which gives $f(x)=\frac{1}{x} \ \blacksquare$ Note:Solution is wrong since continuity of the function is not proven, hence we can't take $y \rightarrow 0$

Because of the original equation there exist an $x_0$ such that $f(x_0)>1$ As solutions above, we can show $f$ is injective. $P(x_0,\frac{f(x_0)-1}{f(x_0)}) $ gives $f(x_0+ \frac{f(x_0)-1}{f(x_0)}) = 1$. So there is an $x_1$ such that $f(x_1)=1$ $P(x,x_1-x),(x<x_1)$ gives $f(x)= \frac{1}{x+1-x_1}$ Plugging infinite $x,y$ pairs such that $x+y<x_1$ and $yx_1>x$ gives $x_1 = 1$ and $f(x)=x $ for $x<1$ $P(1,y)$ gives $f(f(x))=x$ for $x>1$, but $f(\frac{1}{x})=x= f(f(x)) $ and from injectivity $f(x)=\frac{1}{x}$ for $x>1$ Hence, $f(x)= \frac{1}{x}$ for all $x>0$

we note h(χ) the function such that if x>0 so h(x)=f(x) and h(x)=0 if x=0 proof : h(xh(x+y))=yh(x)+1 p(0,y): h(0)=yh(0)+1 so if h(0)≠0 y=1/h(0)+1 which is a point fix a contra diction because y is free over R+ so h(0)=0 p(x,-x);remark h is defind over R so h(0)=-xh(x)+1 then h(x)=1/x so over R+ :f(x)=1/x

Solved with OronSH and pi271828. The only solution is $\boxed{f(x) = \frac{1}{x} } $. To see this works, note the left hand side becomes $\frac{x+y}{x}$ and the right hand side becomes $\frac{y}{x} + 1 = \frac{x+y}{x}$. Now we prove it's the only solution. Claim: $f$ is injective. Proof: Suppose $f(a) = f(b)$ for some positive reals $a,b$. Fix an $0<x < \min(a,b)$. $P(x, a-x): f(x f(a)) = (a-x) f(x) + 1$. $P(x,b-x): f(xf(a)) = (b-x) f(x) + 1$. So $(a-x)f(x) + 1 = (b-x) f(x) + 1\implies a = b$. $\square$ \[P\left( x, \frac{1}{f(x)} \right) : f\left( x f\left(x + \frac{1}{ f(x)} \right)\right) = 2\implies f\left( x + \frac{1}{f(x)} \right) = \frac{k}{x} \ \ \ \ \ \ \ \ (1) \]where $k$ is the unique positive number with $f(k) = 2$. The RHS of $(1)$ takes all positive real numbers, so $f$ is surjective. Now plugging $x = k$ in $(1)$ gives that $f\left( k + \frac{1}{2} \right) = 1$. For $x < k + \frac{1}{2}$, $P\left( x, y = k + \frac{1}{2} - x \right):f(x) = yf(x) + 1 $, so \[f(x) = \frac{1}{1-y} = \frac{1} {x + \frac{1}{2} - k }\]This implies that $f\left( \frac{k}{2} \right) = \frac{2}{1 - k}$. For any $x < k$, $P(x, k - x): f(2x) = (k-x) f(x) + 1$. Setting $x = \frac{k}{2}$ gives that $2 = \frac{k}{2} f\left( \frac{k}{2} \right) + 1,$ so $f\left( \frac{k}{2} \right) = \frac{2}{k}$, which means \[ \frac{2}{k} = \frac{2}{1-k} \implies k = \frac{1}{2} \] Then we have $f(x) = \frac{1}{x}$ for all $x<1$. Now since $f\left( x + \frac{1}{f(x)} \right) = \frac{1/2}{x}$, if $f(x) = \frac{1}{x}$, then $f(2x) = \frac{1}{2x}$, which means if all $x\in (a,b)$ satisfy $f(x) = \frac{1}{x}$, then all $x$ in $(2a, 2b)$ satisfy $f(x) = \frac{1}{x}$. Since all $x\in (0,1)$ satisfy $f(x) = \frac{1}{x}$, we can induct to get that all $x\in (0, 2^n)$ for any nonnegative integer $n$ satisfy $f(x) = \frac{1}{x}$. Since $2^n$ is unbounded, all positive reals $x$ satisfy $f(x) = \frac{1}{x}$.

Nice! Let $P(x, y)$ be the given assertion. Firstly, we claim that the function is injective. Let $f(a) = f(b)$. $P(x, a - x)$, $P(x, b - x)$ gives: $$af(x) + 1 = bf(x) + 1 \implies a = b$$After substituting $P(x, \frac{1}{f(x)})$, we see that the function is surjective. Now, let $f(\alpha) = 1$. $P(1, y)$ yields: $$f(f(y + 1)) = yf(1) + 1$$Plugging in $P(f(y) + 1, \frac{yf(1)}{yf(1) + 1})$, we get: $$f(f(y+1)f(f(y+1) + \frac{yf(1)}{yf(1) + 1})) = \frac{yf(1)}{yf(1) + 1)}\times f(f(y+1)) + 1 = yf(1) + 1$$By injectivity of $f(x)$, we have: $$f(f(y+1)f(f(y+1) + \frac{yf(1)}{yf(1) + 1})) = f(f(y + 1)) \implies f(y+1)f(f(y+1) + \frac{yf(1)}{yf(1) + 1} = f(y + 1)$$$$\implies f(y + 1) + \frac{yf(1)}{yf(1) + 1} = 1$$Since $f(\alpha) = 1$, we have: $$\alpha - \frac{yf(1)}{yf(1) + 1} = f(y + 1) \; \; \; \; (1)$$For sufficiently large $y$, $\frac{yf(1)}{yf(1) + 1}$ tends to $1$ (but not exactly 1), and since $f(y + 1)$ is positive we have $\alpha \geq 1$. We now claim that $f(1) = 1$. Say, for the sake of contradiction, $\alpha > 1$. $P(\alpha - 1, 1)$ gives: $$f(\alpha - 1) = f(\alpha - 1) + 1$$This is clearly not possible which means that $\alpha = 1$. Substituting this value in $(1)$, we have: $$f(y + 1) = \frac{1}{y + 1}$$For $y \geq 1$, we have $f(y) = \frac{1}{y}$. We prove this for $y < 1$ as well. $P(y, 1 - y)$ gives: $$f(y) = (1 - y)f(y) + 1 \implies yf(y) = 1 \implies f(y) = \frac{1}{y}$$Hence, $f(x) = \frac{1}{x}$

ZETA_in_olympiad wrote: Pick an $a$ for $f(a)=1.$ If $a>1$ then $P(a-1,1)$ gives contradiction. If $a<1$ then take $b$ for $f(\tfrac{1}{b})=1-a.$ By $P(\tfrac{1}{b},a)$ we get contradiction. Note that $f(1)=1.$ Can you please clarify how you get contradiction for $a<1$ ?

As usual let $P(x;y)$ denote the given assertion. Assume the function is not injective and let $f(a)=f(b)$ than setting $x+y_1=a$ and $x+y_2=b$ $P(x;y_1)-P(x;y_2) \Rightarrow (a-b)f(x)=0$, a contradiction. Now $P(x;\frac{y}{f(x)}) \Rightarrow f(xf(x+\frac{y}{f(x)}))=y+1$, in particular the expression inside only depends in $y$, not in $x$, so let $xf(x+\frac{y}{f(x)})=g(y)$, and setting $x$ to be $g(y)$ we get that there is a value $\alpha$ such that $f(\alpha)=1$. Take $x+y=\alpha$ and we get that for any $x<\alpha, f(x)=\frac{1}{x+1-\alpha}$ and now we know the form of $f$ for sufficiently small inputs. Fixing $x+y<\alpha$ and taking $x$ very very small we get $\alpha=1$ and now again fixing $x+y$ but now for arbitrary values and taking $x\rightarrow 0$ we get that $f(a)=\frac{1}{a}$ for any positive real $a$ which is indeed a solution.

Wooahhhhhh beautiful and crazy problem First analysis-style functional equation that I’ve solved myself yay This is going to be a LOT different to the solutions above I don’t know how I managed to cook this up. So firstly we claim $f(x)=\frac{1}{x}$ is a working function, which is easy checking. Now we check that the related functions $f(x)=\frac{c}{x}$ for $c\neq1$ don’t work. Indeed: \begin{align*} f(xf(x+y))&=f\left(x\frac{c}{x+y}\right)\\ &=\frac{x+y}{x}\\ &=\frac{y}{x}+1\\ yf(x)+1&=\frac{cy}{x}+1 \end{align*}so we’re forced to have $c=1$. So it suffices now to show that $f(x)=\frac{c}{x}$ for some constant $c$ as per the above. We now split the proof into 5 (!) parts, the first three of which are easy and elementary, whilst the last two really start to come scarily close to real analysis Part 1: $f$ is injective and surjective over all real numbers $>1$ Assume $f(a)=f(b)$. Choose $c<a,b$, and $A=a-c$, $B=b-c$. \begin{align*} f(a)&=f(b)\\ f(c+A)&=f(c+B)\\ f(cf(c+A))&=f(cf(c+B))\\ Af(c)+1&=Bf(c)+1\\ A&=B\\ a&=b \end{align*}Also note by varying $y>0$ that $f(xf(x+y))$ attains any positive real number $>1$. Part 2: $af(a+yf(b))=bf(b+yf(a))$ for any two $a,b\in\mathbb{R}_{>0}$ Notice both are just equal to $f^{-1}(yf(a)f(b)+1)$ (here the inverse is well defined as clearly from the functional equation some $c\in\mathbb{R}_{>0}$ satisfies $f(c)=yf(a)f(b)+1$, and since $f$ is injective this $c$ is unique). Part 3: $f$ strictly decreasing Take $a<b$. Note $f(a)\neq f(b)$ so assume FTSoC $f(a)<f(b)$. Now consider a $y\in\mathbb{R}_{>0}$ such that $a+yf(b)=b+yf(a)$ (this actually can be explicitly computed! $y=\frac{b-a}{f(b)-f(a)}>0$ by assumption). Hence $af(b+yf(a))=af(a+yf(b))=bf(b+yf(a))$, so $a=b$, contradiction. Part 4: $\lim_{x\to\infty}f(x)=0$ Notice that $f(x)$ is a strictly decreasing function. Since it is bounded below by $0$, we can safely claim that $\lim_{x\to\infty}f(x)$ is well behaved! In fact it exists! Call it $A$. First we do some scouting. Notice that since on the RHS $y$ is free to move around we have some degree of freedom. This calls for some application of inequalities \begin{align*} f(xf(x+y))&=yf(x)+1\ge Ay+1\\ xf(x+y)&\ge xA>0\\ f(xf(x+y))&\le f(xA)\\ f(xA)&\ge Ay+1 \end{align*}So if we send $y\to\infty$, the fact that $A>0$ implies $Ay+1\to\infty$. This implies $f(xA)\ge f(xA)+1$ in particular, contradiction. Hence $A=0$ and $\lim_{x\to\infty}f(x)=0$. (Only just realised this part was actually redundant oops keeping this still because 4 is an unlucky number) Part 5: Finishing up with chains and more inequalities First we pick any $a<b$ (!). In the style of APMO 2023/4 let $(y_n)_{n=-\infty}^{\infty}$ be such that $y_0=1$ and $b+f(a)y_{n-1}=a+f(b)y_n$. Let $x_n=a+f(b)y_n$. Note now $af(x_n)=bf(b+f(a)y_n)=bf(a+f(b)y_n)=bf(x_{n+1})$, so $f(x_{n-1})=\frac{b}{a}f(x_n)>f(x_n)$. Hence $x_{i}$ is strictly increasing. Notice that $f(x_{i})=\frac{b^i}{a^i}f(x_0)=\left(\frac{b}{a}\right)^if(x_0)$ so $x_{-i}\to0$ as $i\to\infty$. The reason for this is as follows: first notice that from $f$ surjective on real numbers $>1$ that $\exists c\in\mathbb{R}_{>0}$ such that $f(c)=r$ for any $r>1$. Now since $(x_{-n})_{n=0}^{\infty}$ is strictly decreasing and bounded from below by $0$, $c=\lim_{n\to\infty}x_{-n}$ exists again by the monotone convergence theorem. Hence noticing that if $x_{-i}\to c$ for some $c>0$, there are still $0<d<c$, so considering $f(d)$, $f(x_{-i})<f(d)$, hence $\left(\frac{b}{a}\right)^nf(x_0)<f(d)$ for all $n>0$, a clear contradiction since $\frac{b}{a}>1$. Hence $\lim_{n\to\infty}x_{-n}=0$. But then this implies quite quickly that $\lim_{n\to\infty}y_{-n}=-\frac{a}{f(b)}$. However we can compute $y_{-n}$ quite explicitly. Noticing that $y_{n-1}=\frac{a-b+f(b)y_n}{f(a)}$, we can quite quickly deduce from induction that for $n\ge1$: \begin{align*} y_{-n}&=\frac{a-b}{f(a)}\left(\sum_{i=0}^{n-1}\left(\frac{f(b)}{f(a)}\right)^i\right)+\frac{f(b)^{n}}{f(a)^{n}} \end{align*}However notice in the limit as $n\to\infty$, $\frac{f(b)^{n}}{f(a)^{n}}\to0$, whilst since $f(b)<f(a)$: \begin{align*} \frac{a-b}{f(a)}\left(\sum_{i=0}^{n-1}\left(\frac{f(b)}{f(a)}\right)^i\right)&\to\frac{a-b}{f(a)}\left(\sum_{i=0}^{\infty}\left(\frac{f(b)}{f(a)}\right)^i\right)\\ &=\frac{a-b}{f(a)}\frac{1}{1-\frac{f(b)}{f(a)}}\\ &=\frac{a-b}{f(a)-f(b)} \end{align*}So $\lim_{n\to\infty}y_{-n}=\frac{a-b}{f(a)-f(b)}=-\frac{a}{f(b)}$. Now by a very useful result called addendo, $\frac{a-b-a}{f(a)-f(b)-(-f(b))}=-\frac{a}{f(b)}$, or $\frac{b}{f(a)}=\frac{a}{f(b)}$. Hence $af(a)=bf(b)$. We never really fixed down $a$ and $b$ though, so in fact $af(a)$ is constant which is to say $f(a)=\frac{c}{a}$ for some constant $c$ for all real numbers $a>0$, ending the proof.

\[f(xf(x + y)) = yf(x) + 1\]Only function satisfying the equation is $f(x)=\frac{1}{x}$. Claim: $f$ is injective. Proof: Suppose that $f(a)=c=f(b)$. Pick $y=a-x$ and then $y=b-x$ (and choose $x$ sufficiently small) in order to see that $(b-x)f(x)+1=f(cx)=(a-x)f(x)+1$ which implies $a=b$ so $f$ is injective.$\square$ Claim: $f$ is surjective. Proof: Plugging $x,y/f(x)$ gives $f$ is surjective for $>1$. \[f(xf(x+\frac{y}{f(x)}))=y+1=f(zf(z+\frac{y}{f(z)}))\implies xf(x+\frac{y}{f(x)})=zf(z+\frac{y}{f(z)})\]$z=1$ yields $xf(x+\frac{y}{f(x)})=f(1+\frac{y}{f(1)})$. Write $xf(1+\frac{y}{f(1)})$ instead of $x$ to see that $f$ is surjective.$\square$ Claim: $f(1)=1$. Proof: Let $f(a)=1$ and suppose that $a\neq 1$. For $x<a$, we have $f(x)=f(xf(a))=(a-x)f(x)+1$ or $f(x)=\frac{1}{x+1-a}$. Note that this requires $1\geq a$ hence $1>a$ by our assumption. Pick $x<a(1-a)<a$ and $x+y<a$ to get $f(\frac{x}{x+y+1-a})=\frac{y}{x+1-a}+1=\frac{x+y+1-a}{x+1-a}$. Since $\frac{x}{x+y+1-a}<a$ holds, we see that $\frac{1}{\frac{x}{x+y+1-a}+1-a}=\frac{x+y+1-a}{x+1-a}$ or $x+(1-a)(x+y+1-a)=x+1-a$ or $(1-a)(x+y-a)=0$ which is impossible.$\square$ Since $f(1)=1$, we have $f(x)=\frac{1}{x}$ for $x<1$. This implies $f(x)>1\iff x<1$ by bijectivity. We have $f(xf(x+y))>1$ so $xf(x+y)<1$. Thus for $x<1$, \[\frac{1}{xf(x+y)}=\frac{y}{x}+1=\frac{x+y}{x}\implies f(x+y)=\frac{1}{x+y}\]As desired.$\blacksquare$