Claim 1: $f$ is increasing Proof. FTSOC let there exist $x_1<x_2, f(x_1)>f(x_2)$. Then taking $y$ s.t. $x_1+yf(x_1)=x_2+yf(x_2)$ and comparing $P(x_1,y), P(x_2,y)$ we obtain a contradiction. Hence all limits do exist from above. By $lim_{x\rightarrow x_0}f(x)$ we will be referring exactly to this limit from above. Claim 2: $f(1)\geq 1$ Proof. FTSOC let $c=f(1)<1$. Then $P(1,y)$ gives us $f(1+yc)=f(1+y)$ or $f(z)=f(1+\frac{z-1}{c})$ for $z>1$. Defining $g(x)=1+\frac{x-1}{c}$ we get that $f(z)=f(g^n(z))$, which would imply from Claim 1 that $f(x)$ is constant for every $x>z$. However, fixing an $x\neq 1$ and taking a large enough $y$ we get a contradiction. Claim 3: $\lim_{x\rightarrow 1}f(x)=1$ Proof. Let $\lim_{x\rightarrow 1}f(x)=t, \lim_{x\rightarrow ty+1}f(x)=k$ for a fixed $y$. Taking $x\rightarrow 1$ in the original equation we get: $$k=\lim_{x\rightarrow 1}(xf(y+1))=f(y+1) \Rightarrow \lim_{x\rightarrow ty+1}f(x)=f(y+1)\geq f(ty+1) \Rightarrow 1\geq t$$In the last inequality if $1<t$ we would get a contradiction similar to the one in Claim 2. However we also have $lim_{x\rightarrow 1}f(x)=t\geq f(1)\geq 1 \Rightarrow \lim_{x\rightarrow 1}f(x)=1$ Claim 4: $\lim_{x\rightarrow x_0}f(x)=x_0$ Proof. Taking $y\rightarrow 0$ int the original FE. Claim 5: $f(x)=x, x>1$ Proof. Taking $x\rightarrow 1$ we get $f(y+1)=\lim_{x\rightarrow 1}f(f(x)y+x)=y+1.$ Claim 6: $f(x)=x$ Proof. Fixing $x$ and taking a large enough $y$ in the original equation we get $\boxed{f\equiv id}$, which is indeed a solution.

We only bound. Claim 1 : $f(x)\geqslant x$ for all $x$. Proof : Let $t>1$. Suppose possible $f(t)<1$. Then $P\left(t,\frac{t-1}{1-f(t)}\right)$ implies $t=1$ which contradicts our assumption. Hence $t>1 \implies f(t)\geqslant 1$. $P(x,y) : f(x+yf(x)) = xf(1+y)\geqslant x$. Set $z=x+yf(x)$ and hence for any $z>x$ we have $f(z)\geqslant x$. Letting $x\rightarrow z^{-}$ we infer that $f(z)\geqslant z$ for any $z$. $\blacksquare$ Claim 2 : $f(x) \geqslant 1+c(x-1)$ for all $x$, where $c=\sup \frac{f(x)}{x}$. Proof : $P(x,y) : xf(1+y) = f(x+yf(x))\geqslant x+yf(x)$. Hence $f(1+y) \geqslant 1+\left(\frac{f(x)}{x}\right)y$. Note $\frac{f(x)}{x}$ must be upper bounded by last equation. Hence $c\in \mathbb{R}$. For any $\varepsilon>0$, choose $x$ such that $\frac{f(x)}{x}\geqslant c-\varepsilon$. Hence from last equation, $f(1+y)\geqslant 1+cy - \varepsilon y$. Letting $\varepsilon\rightarrow 0$, we infer that $f(1+y)\geqslant 1+cy \text{ for all } y$. Hence we have proven the claim for $x>1$. For $x<1$, it trivially holds since $f(x)\geqslant x\geqslant 1+c(x-1)$ with the last inequality holding since $c\geqslant 1$. $\blacksquare$ Claim 3 : $f(x)=x$ for all $x$. Proof : $P(x,y) : xf(1+y) =f(x+yf(x))\geqslant 1+c(x+yf(x))-1) \geqslant 1+c(x+yx-1)$. Hence $f(1+y) \geqslant c(1+y)+\frac{1-c}{x}$. Letting $x\rightarrow \infty$ we infer that $f(1+y)\geqslant c(1+y)$ for any $y$. But recollect $c=\sup \frac{f(x)}{x}$. Hence $f(z)=cz$ for all $z>1$. But from $P(x,y)$ where $x>1$, infer that $c=1$. Hence $\frac{f(x)}{x}\leqslant c=1$. Combining with claim 1, we infer $f(x)=x$ and we are done! $\blacksquare$

Claim 1. : $f(x)\geq 1$ for $x>1$ Suppose there is $x>1, f(x)<1$. Take $y=\frac{1-x}{f(x)-1}$. Then $$f(x+yf(x))=f(1+y)$$so $x=1$ which contradicts $x>1$. Claim 2.: $f(x)\geq x$ Let $\epsilon > 0$. Set $y=\frac{\epsilon}{f(x)}$ $$f(x+\epsilon)=xf(1+y)\geq x$$As we can take $\epsilon$ as small as we want we proved $f(x)\geq x$ Claim 3.: There exists $C$ such that $f(x)\leq Cx$ $$xf(1+y)=f(x+yf(x))\geq x+yf(x)$$$$f(1+y)\geq 1+y \frac{f(x)}{x}$$When we fix $y$ then $\frac{f(x)}{x}$ is bounded from above. Out of all upper bounds we take infimum for $C$. We know $f(x)\geq x$ therefore $C \geq 1$ Claim 4. : There exists $k$ such that $f(x)\geq Cx-k$ for $x>1$ Let $\epsilon \geq 0$. Because of definition of $C$ there exists $a$ such that $f(a)\geq (C-\epsilon)a$ We will set $x=a$ in inequality we got in last claim. $$f(1+y)\geq 1+y \frac{f(a)}{a} \geq 1 +(C- \epsilon)(y+1) - (C - \epsilon)$$This means for all $x>1$ $$Cx\geq f(x) \geq (C-\epsilon)x + 1-C+\epsilon$$By taking $\epsilon$ arbitrarily small we get wanted claim. Claim 5. : $f(x)=x$ $$Cx(1+y)\geq xf(1+y) = f(x+yf(x)) \geq Cx+Cyf(x)-k \geq Cx + C^2 yx-Cky-k$$$$(C-C^2)xy \geq -Cky-k$$Let's note that $C-C^2 \leq 0$. If it is nonzero (negative) we can take arbitrarily large $x$ that makes a LHS arbitrarily large negative number while RHS is fixed. That's contradiction so $C=1$. $x\geq f(x)\geq x$ which finishes the proof.

I think you can use the fact that any surjective, increasing function must be continuous and then take $y\rightarrow 0^{+}$.

sami1618 wrote: I think you can use the fact that any surjective, increasing function must be continuous and then take $y\rightarrow 0^{+}$. I might be wrong, but doesn't it have to be strictly increasing to show it must be continuous?

$f$ is increasing. Suppose exist $x_1<x_2$ s.t. $f(x_1)>f(x_2)$. The equation $x_1+yf(x_1)=x_2+yf(x_2)$ has a solution for some $y>0$. So $x_1f(y+1)=f(x_1+yf(x_1))=f(x_2+yf(x_2))=x_2f(y+1)$ Contradiction so $f$ is increasing. $f$ is surjective because fixing $y$ and moving $x$ the $RHS$ can be every positive real number. $f$ surjective and increasing means that $f$ is continuous. So take $y\to 0^+$ $f(x)=xf(1)$ Sostitute in the equation gives that $f(x)=x$ is the only solution.

Denote $P(x,y)$ the assertion of the given F.E. Claim 1: $f$ is an increasing function. Proof: FTSOC there is $a>b$ with $f(b)>f(a)$ then from $P \left(a, \frac{a-b}{f(b)-f(a)} \right)$ and $P \left(b, \frac{a-b}{f(b)-f(a)} \right)$ we get they share same LHS but RHS is $af(\text{something})=bf(\text{something})$ so $a=b$, contradiction. Claim 2: $f$ is surjective. Proof: Just fix $y$ and vary $x$ to get it. Claim 3: $f(1)=1$. Proof: First note that when $x>1$ and $f(x)<1$ replacing some $y$ gives $x=1$ so if $x>1$ then $f(x) \ge 1$, and in the same way if $x<1$ then $f(x) \le 1$ so we get $f(x+y) \ge x$ for all positive reals $x,y$ (in fact by letting $x \to x-y$ and then $y \to 0$ we can get $f(x) \ge x$ for all $x \in \mathbb R^+$), and also from $P(1,x)$ we get $f(1+xf(1))=f(1+x)$ which inducting gives $f(1+xf(1)^n)=f(1+x)$ for all integers $n$, now trivially if $f(1)>1$ then $f(1+x) \ge xf(1)^n$ and making $n \to \infty$ gives contradiction, same way if $f(1)<1$ then $f(1+x) \ge xf(1)^n$ and $n \to -\infty$ gives a contradiction, therefore $f(1)=1$ as desired. Claim 4: There exists $C$ such that $f(x) \le Cx$ for all $x \in \mathbb R^+$ Proof: $xf(1+y) \ge f(x)$ so fixing $y$ such that $f(1+y)$ reaches an infimun among all gives our $C$ (so we let $C$ be the supremun), in addition we can go stronger and say $xf(1+y) \ge x+yf(x)$ from the observation made in the previous claim and set $x \to 0$ here for some fixed $y$ to get that $0 \ge y \cdot \lim_{x \to 0} f(x)$ so in fact $\lim_{x \to 0} f(x)=0$. The finish:Note that $f(1+y) \ge 1+y \cdot \frac{f(x)}{x}$ so dividing by $1+y$ and taking $y \to \infty$ gives that $\lim_{z \to \infty} \frac{f(z)}{z} \ge \frac{f(x)}{x}$ for all positive reals $x$, now pick $x$ extremely close to the supremun (as needed) to get that in fact $\lim_{z \to \infty} \frac{f(z)}{z}=C$ because if it was remotely less it is attainable and even suprassed by some $x$ and obviously this can be repeated for any sequence that increases indefinitely eventually thanks to increasingness. Now on $P(x,y)$ we get: \[ \frac{f(x+yf(x))}{x+yf(x)}=\frac{x(y+1)}{x+yf(x)} \cdot \frac{f(y+1)}{y+1} \overset{y \to \infty}{\implies} C=\frac{x}{f(x)} \cdot C \implies f(x)=x \]Therefore $f(x)=x$ for all positive reals $x,y$ thus we are done .

Increasing + surjective = continuous, set $y \to 0$.

A longer, but quite unique solution:

$P(1,x) : f(1+xf(1)) = xf(1+x)$ Claim 1: $f(1) = 1$ Proof: assume $f(1) \neq 1$. Assume there exists a $c$ s.t. $f(c) = 1$ (obv $c \neq 1$ and there does exist a $c$ because the function is surjective by changing the value of x) $P(c,x) : f(c+x) = cf(1+x)$ Now since the difference between $1+xf(1)$ and $1+x$ can be any value from 0 (not including 0) to infinity, while the difference between $c+x$ and $1+x$ is constant, we can say that there is a value $d$ s.t. $f(c+d) = f(1+d)$. $P(c,d) : f(c+d) = cf(1+d) \Rightarrow 1=c$ Therefore, $f(1) = 1$ Claim 2: $f(x)$ is increasing Proof: $P(x,k) : f(x+kf(x)) = xf(1+k)$ Suppose there exist $a, b$ s.t. $a > b$ and $f(a) < f(b)$. If $k = 0$, we have $a+kf(a) > b+kf(b)$, while if $k$ goes to infinity, we have $a+kf(a) < b+kf(b)$, thus by intermediate value theorem, there exist $k$ s.t. $a+kf(a) = b+kf(b)$. However then we have $f(1+k) = \frac {f(a+kf(a))}{a}$ and $f(1+k) = \frac {f(b+kf(b))}{b}$, but $\frac{1}{a} \neq \frac {1}{b}$, thus we have a contradiction. Therefore, this function is increasing. Now we can finish off the solution after 1 more substitution. $P(x, \frac {k}{f(x)}) : f(x+k) = xf(1+ \frac {k}{f(x)})$ Notice that $\lim_{k \to 0} xf(1+ \frac {k}{f(x)}) = x$, thus $\lim_{k \to 0} f(x+k) = f(x) = x$ (The function is surjective and increasing, thus continuous and therefore we can take limits) Thus we are done and the check is trivial.

GeorgeRP wrote: Claim 3: $\lim_{x\rightarrow 1}f(x)=1$ Proof. Let $\lim_{x\rightarrow 1}f(x)=t, \lim_{x\rightarrow ty+1}f(x)=k$ for a fixed $y$. Taking $x\rightarrow 1$ in the original equation we get: $$k=\lim_{x\rightarrow 1}(xf(y+1))=f(y+1) \Rightarrow \lim_{x\rightarrow ty+1}f(x)=f(y+1)\geq f(ty+1) \Rightarrow 1\geq t$$In the last inequality if $1<t$ we would get a contradiction similar to the one in Claim 2. However we also have $lim_{x\rightarrow 1}f(x)=t\geq f(1)\geq 1 \Rightarrow \lim_{x\rightarrow 1}f(x)=1$ Beware that those limits do not necessarily exist, so you cannot work with them like that. Since the function is monotone, it only follows that lateral limits exist in each point, but not necessarily limits. One way to fix this is by noticing that $f$ is surjective. This combined with monotonicity implies continuity, and then your arguments work.

@above In my solution I am referring everywhere to the limit from above. Maybe it wasn't clear with the notaion I was using, so I fixed that now.

This problem was proposed by Ioannis Galamatis (Greece)

I have a unique approach: we investigate what is $f(>x)$ for reals $x$ and make deductions: Consider what value can LHS and RHS attain when we fix some real $x$ and change $y$ freely. The LHS values are $f(>x)$, and RHS - $xf(>1)$. Thus, this two sets are identical. Suppose $t \in f(>1)$. Then for $x>1$, $xt \in f(>x)$, but $f(>x)$ is a subset of $f(>1)$, hence $xt \in f(>1)$. So $[t,+\infty ) \in f(>1)$. We have two cases: Case 1: $f(>1)=[t,+\infty)$ for some real t. Then $f(>x)=[tx,+\infty)$. Let $a$ be a number($a>x$) such that $f(a)=tx$. Then for any $b<a$ we know $tx \in f(b)=[tb,+\infty )$, hence $b \leq x$. But obviously it is false for $b=\frac{a+x}{2}$, contradiction. Case 2: $f(>1)=(t,+\infty)$ for some real t. We can change $x$ freely in RHS, hence our function is surjective. Let's try setting $y=\frac{x-1}{1-f(x)}$. If you can, then $x=1$, but $y$ must be positive, so you can't. Then for all $x \neq 1$ either $f(x)=1$ or $f(x)$ and $x$ lie on different sides of $1$. So for all $x>1$ $f(x) \geq 1$ and for all $x<1$ $f(x) \leq 1$. But then as $f(>1)=(t,+\infty)$ we must have $t=1$, and so $f(>x)=(x,\infty )$ for all positive reals $x$. Also from this for all $x$ $f(x) \geq x$, and so for $x>1$ $f(x) \neq 1$. Suppose for some $x<1$ equality $f(x)=1$ holds. Then $$f(x+y)=xf(1+y)$$which rewrites as $$f(y+(1-x))=f(y) \frac{1}{x}$$for $y>x$. But then $$f(y+(1-x)k)=f(y) (\frac{1}{x})^k$$Note that if we take big $y+(1-x)k$ so that $f(y+(1-x)k)$ is close to $y+(1-x)k$, then $f(y) \geq y$ is approximately $(y+(1-x)k)x^k$. When we take $k$ big enough it will become arbitrarily close to $0$, so the inequality $f(y) \geq y$ won't hold. Hence, for all $x \neq 1$ $f(x) \neq 1$, and so $f(1)=1$. Now suppose there exist $a>b>1$ with $f(a)=f(b)$. Then setting $x=\frac{1}{f(a)}$, $y=a-1$ and $y=b-1$ yields a contradiction. Hence the function is injective on $x \geq 1$. But then for all $x \geq 1$ we must have $f(x)=x$. Now we return to changing $y$ freely. The RHS attains all values once, hence also the LHS, hence $f$ is injective on all positive reals. Then $f(x)=x$ for all positive $x$. It is easy to check that this function satisfies the condition.

Denote $P(x,y)$ the assertion of the F.E. Claim 1: $f$ is an increasing function. Proof: Assume otherwise, there is $a>b$ with $f(b)>f(a)$ then there exists $y>0$ such that: $a+yf(a)=b+yf(b)$ which implies $a=b$, a contradiction. Claim 2: $f$ is injective. Proof: If $f(z)=f(t)$ with $z>t$, then: $P(z,y)$ and $P(t,y)$ give: * $f(z+yf(z))=zf(y+1)$ * $f(t+yf(z))=tf(y+1)$ So that $f(z+yf(z))=\frac{z}{t} f(t+yf(z))$ Which can be rewritten as: $f((z-t)+x)=\frac{z}{t}f(x)$, for $x>t$. $\quad(*)$. Now, if $a>b$ and $f(a)=f(b)$, then we must have $f(a)=f(\frac{a+b}{2})=f(b)$, since the function is increasing. Using $(*)$ with $z=a,t=\frac{a+b}{2}$ then with $z=\frac{a+b}{2}, t=b$ gives: $\frac{a}{\frac{a+b}{2}}=\frac{\frac{a+b}{2}}{b} \implies a=b$ Claim 3: $f(x)=x, \forall x \neq 1$: Proof: $P(\frac{1}{y},y): f(\frac{1}{y}+yf(\frac{1}{y}))=\frac{1}{y}f(y+1)$ $P(y, \frac{1}{f(y)}): f(y+1)=yf(\frac{1}{f(y)}+1)$ Combining these two and using injectivity, we get: $\frac{1}{y}+yf(\frac{1}{y})=\frac{1}{f(y)}+1$ And, by symmetry: $y+\frac{1}{y}f(y)=\frac{1}{f(\frac{1}{y})}+1$ Solving this for $y \neq 1$ (steps omitted cuz lazy), we get $f(y)=y$. And we can easily extend this to $y=1$, so $f(x)=x$.

Here is my solution, sorry that I didn’ t use LaTeX Call this equation P(x,y) =f(x+y.f(x))=x.f(y+1) P(x,1) give us f(x+f(x))=f(2).x, call f(2)=C P((x+f(x)), 1): f(x+f(x)+C.x)=C.(x+f(x)) P(x, (f(x)+C.x)/f(x)): f(x+C.x+f(x))=x.f((C.x+f(x))/f(x) + 1) We get that C.(x+f(x))=x.f((C.x+f(x))/f(x) + 1) <=> C.(f(x)/x + 1)=f(2+ C.x/f(x)), but the right hand side is equivalent to f(2+C.x/f(x))=2.f(x/f(x) + 1), because P(2, x/f(x)) Now C.(f(x)/x + 1)=2.f(x/f(x) + 1) Now we will prove surgectivity and increasing of the function: Claim 1: f(x) is surgective: Prove: let a be fix number, then P(y/f(a+1), a) give us f(y/f(a+1) + a.f(y/f(a+1)))=y, which means surgective from f(k)=y for some y Claim 2: f is increasing: Prove: let x>y and f(y)>f(x), for the sake of contradiction, let P(x, (x-y)/f(y)-f(x)) and P(y, (x-y)/f(y)-f(x)), comparing this two will give us x=y Contradiction, i.e x>y, f(x)>f(y), and also this means injection Now I want to say that this two claims become from solutions of @above, just use the same two, because I would do the same thing to prove this two Now we have surgectivity, let f(k)=x, for k#x Now in this: (C/2).(f(x)/x + 1) = f(x/f(x) + 1), let x->k, then (C/2).(x/k + 1)=f(k/x + 1) Now x->k.x (C/2).(x+1)=f(1/x + 1) x->1/x, (C/2).(1/x + 1) = f(x+1) X-> x-1 (C/2).(x/x-1)=f(x), put this in original equation, we will get f(x)=(2.x^2-3.x)/(1-2.x), which is not true for x>3/2, contradiction with R+, i.e we must have f(x)=x

Fix $y$. Then $g(x)=xf(1+y)$ therefore $Im(g)=\mathbb{R}$. Now let $x_{1} \neq x_{2}$ be positive. Fix $y$ s.t. $x_{1} + yf( x_{1} ) = x_{2} + yf( x_{2} )$ $y=-\frac{\Delta f}{\Delta x}$ If $f$ is non-increasing then $ x_{1} = x_{2} $. Contradiction. $f$ is increasing. $f$ is increasing and surjective therefore it is continuous. $f(x)=\lim_{y\to0^+} f(x+yf(x))=\lim_{y\to0^+} xf(1+y)=xf(1)=cx$ $f(x)=cx$ $cx+c²xy=cx+cxy$ $c²=c$ $c=1$ or $c=0$ If $c=0$, then $f=0$ contradiction. If $c=1$, then $f=id$.

Let the assertion in the OP be $P(x,y) =f(x+yf(x))=xf(y+1)$. $P(x,1)$ gives us $f(x+f(x))=f(2)x=Cx$ where C=$f(2)$. $P((x+f(x)), 1): f(x+f(x)+Cx)=C(x+f(x))$ $P(x, \frac{f(x)+Cx}{f(x)}): f(x+Cx+f(x))=xf(\frac{Cx+f(x)}{f(x)} + 1)$ We get that $C(x+f(x))=xf(\frac{C.x+f(x)}{f(x)} + 1) \iff$ $C(\frac{f(x)}{x} + 1)=f(2+ \frac{Cx}{f(x)})$, but the right hand side is equivalent to $f(2+\frac{Cx}/{f(x)})=2f(\frac{x}/{f(x)} + 1)$, because $P(2, \frac{x}{f(x)})$ Now $C(\frac{f(x)}{x} + 1)=2f(\frac{x}{f(x)} + 1)$ Now we will prove surgectivity and increasing of the function: Claim 1: $f(x)$ is surgective: Proof: let $a$ be a fixed number, then $P(\frac{y}{f(a+1)}, a)$ gives us $f(\frac{y}{f(a+1)} + af(\frac{y}{f(a+1)})=y$, which means surgective from $f(k)=y$ for some $y$ Claim 2: $f$ is increasing: Prove: let $x>y$ and $f(y)>f(x)$, for the sake of contradiction, let $P(x, \frac{x-y}{f(y)-f(x)})$ and $P(y, \frac{x-y}{f(y)-f(x)})$, comparing this two will gives us $x=y$ Contradiction, i.e $f$ is increasing, and also this means injection Now I want to say that these two claims come from solutions of @above, just use the same two, because I would do the same thing to prove these two Now we have surgectivity, let $f(k)=x$, for $k \neq x$ Now in this: $\frac{C}{2}(\frac{f(x)}{x} + 1) = f(\frac{x}{f(x)} + 1)$, let x \to k, then $\frac{C}{2}(\frac{x}{k} + 1)=f(\frac{k}{x} + 1)$ Now x \to kx $\frac{C}{2}(x+1)=f(\frac{1}{x} + 1)$ $x \to \frac{1}{x}$, $\frac{C}{2}(\frac{1}{x} + 1) = f(x+1)$ $x \to x-1$ $\frac{C}{2}\frac{x}{x-1}=f(x)$, put this in original equation, we will get $f(x)=\frac{2x^2-3x}{1-2x}$, which is not true for $x>\frac{3}{2}$, contradiction with $\mathbb{R}^+$, i.e we must have $f(x)$≡$x$.

[REDACTED] fakesolve

Pick a random $c>0$ and let $a=f(c)$. We have $f(c+ax)=cf(1+x)$. Now $x\mapsto c+ax$ yields \begin{align*} (c+ax)f(1+y)=f(c+ax+yf(c+ax))&=f(c+ax+cyf(1+x))\\ &=cf\left(1+x+\frac{cyf(1+x)}{a}\right)\\ &=c(1+x)f\left(1+\frac{cy}{a}\right)\\ &=(1+x)f(c+cy) \end{align*} Since this is true for all $x$, fixing $y$ and matching coefficients gives $f(c+cy)=cf(1+y)$ so $a=c$, or $f(c)=c$.