Hint: change $x$ by $y$ then substract.

Hint

The following solution is strongly inspired by the one the person that proposed this problem showed me. First, we'll prove \[ f(x)\geq x \text{ }(1). \] For fixed $x$, took any $u>f(x)$. Then, for $b = \frac{u-f(x)}{x}$ \[ f(u) = f(f(x)f(b))+x > x \]\[ \Rightarrow f(u) > x, \forall x \in \mathbb{R}_{>0} \] So, if exists $x>f(x)$, then we can substitute $u=x$ in the previous inequality and get $f(x)>x$, which contradicts $x>f(x)$. This implies on the original equation \[ f(xy+f(x) \geq f(x)f(y) + x \text{ }(2) \] Let us prove that $x<1 \Rightarrow f(x)<1$. For $a<1$, took $b=\frac{f(x)}{1-x}$ in $(2)$ \[ f(b) \geq f(a)f(b) + a \Rightarrow f(a) \leq 1 - \frac{a}{f(b)} < 1. \] Now, we'll prove that for every $x<1$, if $f(x)\neq x$, then $f(x)\geq 1-x$. Took inequality $(2)$ in the form \[ f(x) \leq \frac{f(xy+f(x))-x}{f(y)} \leq \frac{f(xy+f(x))-x}{y} \] For a fixed $x<1$, took any $y$ such that $xy+f(x)<1$. Then \[ f(x) < \frac{1-x}{y} \Rightarrow y < \frac{1-x}{f(x)}. \] This shows that for every value of $y<\frac{1-f(x)}{x}$, we have $y<\frac{1-x}{f(x)}$. Which means that doesn't exists $y$ such that $\frac{1-x}{f(x)}<y<\frac{1-f(x)}{x}$. This implies \[ \frac{1-x}{f(x)} \geq \frac{1-f(x)}{x} \] \[ (f(x)-x)(f(x)+x-1)\geq 0 \] And the desired result follows from this inequality. Let $M=\min{\left \{\frac{1}{2}, \frac{1-f(2/3)}{2/3} \right \}}$. Now we'll prove that for every $x\in (0, M), f(x)=x$. Suppose for sake of contradiction that it's false. Then took $u\in (0, M)$ such that $f(u)\neq u$. This implies $f(u)\geq 1-u$. Take $x=2/3, y=u$ in $(2)$. As $u<M$, we have $(2/3)u+f(2/3)<(2/3)M+f(2/3)\leq 1$. Then \[ 1>f((2/3)u+f(2/3))\geq f(2/3)f(u)+2/3 \]\[ 1/3 > (1-u)f(2/3)\geq f(2/3)(1/2) \Rightarrow f(2/3)<2/3 \] which is absurd as $f(2/3)\geq 2/3$. Now we can finally prove that the only solution is $f(x)=x$. First, take fixed points $x$ and $y<1$. The original equation for these values give us \[ f(x(y+1))=f(xy)+x=x(y+1) \] So $x(y+1)$ is also a fixed point. Suppose that exists a maximal $N$ such that for every $x\in(0, N)$, $f(x)=x$ - the last fact showed that exists such $N$ that makes this interval non-trivial. But observe that for every $u\in (N, (1+\min{\{ 1, N \}})N)$ - which is a non empty interval - we also have $f(u)=u$. First, $f(N)=N$ - just took $n$ sufficiently big such that $0<\frac{1}{2^nN-1}<\min{1, N}$ and then took $x=N-\frac{1}{2^n}$ and $y=\frac{1}{2^nN-1}$. Now, took $x=N$ and $y=\frac{u-N}{N}$ - which satisfies $0<y<1$ - in the previous result and we have $f(u)=u$. So, we got an interval $(0, (1+\min{\{ 1, N \}})N)$ such that every point inside it is fixed, and it's stricly bigger then the previous one, which contradicts the hypothesis of $N$ maximal. So, as such $N$ doesn't exists, we have that every point in $(0, + \infty)$ is fixed.

Is there a solution that uses the injectivity of $f$?

Cycle wrote: Is there a solution that uses the injectivity of $f$? Yes, f(xy+f(x)) =f(f(x)f(y)) + x f(xy+f(y))=f(f(x)f(y)) + y Subtracting with y=0 f(f(x))=x+f(f(0)) Now f is bijective. Plugging y=0 again f(f(x)) =x+f(f(x)f(0)) Hence f(0)=0 because f is bijective and f(f(0))=f(f(x)f(0)) f(f(x)) =x Put x=c such that f(c) =1, and y:=y-c f(cy+1-c^2)=y=f(f(y)) cy+1-c^2=f(y) f(x) = cx+1-c^2 It is easy to check that c=-+1.

Krm wrote: Subtracting with y=0 You can not since $0$ does not belong to the domain (positive reals)

Oh sorry. Thank you anyway

We claim that the only solution is $f \equiv x$, which is trivially a solution. Notice that if $x > f(x)$, we know from $P(x, \frac{x - f(x)}{x})$ that $f(x) > x$, which is clearly absurd. Hence, we must have that $$f(x) \ge x, \forall x \in \mathbb{R}.$$ At this point, we may assume that $f(t) > t$ for some $t \in \mathbb{R}$, as otherwise we arrive at our solution $f \equiv x.$ We know that for any $\epsilon > 0$, we have by $P(t, \epsilon)$ that: $$t \epsilon + f(t) \le f(t \epsilon + f(x)) = f(f(t)f(\epsilon)) + t,$$ i.e. $$f(f(t)f(\epsilon)) > f(t) - t.$$ Let $f(t) - t = c$ and $f(t) = d.$ Lemma 1. There is some real $x$ so that $f(x) \ge 3x.$ Proof. Select an arbitrarily small $\epsilon > 0$, so that $9 \epsilon d < c.$ We have $f(f( \epsilon) d) > c.$ If $f(\epsilon) > 3\epsilon$, we're done. Else we have $f(\epsilon) d < \frac{c}{3}.$ But then we have $f(f(\epsilon)d) > 3 f(\epsilon) d,$ so we're again done. $\blacksquare$ Lemma 2. We have that $f(x) > 2x$ for sufficiently large $x.$ Proof. Let $a$ be a real number so that $f(a) \ge 3a$, which exists by Lemma $1.$ We have that $f(ay + f(a)) = f(f(a)f(y)) + a \ge 3ay + a$. For $y$ sufficiently large, we have $f(ay + f(a)) > 2(ay + f(a)),$ and so the lemma is proven. $\blacksquare$ Suppose that there is a constant $C > 10^{10^{10^{10}}}$ so that for all $x > C$, we have $f(x) > 2x.$ Fix a constant $\ell = C+1.$ Claim. For all real numbers $n > f(\ell) + C \ell$, there exists some $m > n$ with $f(m) = f(n) - \ell.$ Proof. Let $y = \frac{n - f(\ell)}{\ell}.$ By $P(\ell, y)$, we have that $f(f(\ell)f(y)) = f(n) - \ell$, so we just need to show that $f(\ell)f(y) > n,$ i.e. $f(\ell)f(y) > f(\ell) + \ell y.$ We have that $$f(\ell)( f(y) - 1) > f(\ell)(2y -1) > f(\ell) y > \ell y,$$ so this is clearly true. $\blacksquare$ Select some arbitrary real $n > f(\ell) + C \ell.$ Let $f(n) = k$. Then, by the Claim applied $\left \lceil k \right \rceil$ times, we know that there is some real number $t > n$ so that $f(t) = k - \left \lceil k \right \rceil \cdot \ell$. However this is negative, absurd. Hence, there is no solution where $f(x) > x$ for some $x \in \mathbb{R}$, and so the only solution is $f \equiv x$. $\square$

Pathological wrote: Proof of Lemma 1. ... Else we have $f(\epsilon) d < \frac{c}{3}.$ ... Why $f(\epsilon)f(t) < \frac{f(t) - t}{3}$? Pathological wrote: Proof of Lemma 2. ... We have that $f(ay + f(a)) = f(f(a)f(y)) + a \ge 3ay + a$. For $y$ sufficiently large, we have $f(ay + f(a)) > 2(ay + f(a)),$ and so the lemma is proven. How do we have $f(ay + f(a)) > 2(ay + f(a))$?

ZeusDM wrote: Why $f(\epsilon)f(t) < \frac{f(t) - t}{3}$? We selected $\epsilon$ so that $9 \epsilon d < c$. If $f(\epsilon) \le 3 \epsilon$, that means that $3f(\epsilon) d \le 9 \epsilon d < c$, so $f(\epsilon)d < \frac{c}{3}.$ ZeusDM wrote: How do we have $f(ay + f(a)) > 2(ay + f(a))$? For sufficiently large $y$, we have that $3ay+a > 2(ay + f(a)).$ Since $f(ay + f(a)) \ge 3ay + a$ for all $y \in \mathbb{R}$, this means that $f(ay + f(a)) \ge 3ay + a > 2(ay + f(a))$ for sufficiently large $y.$

Assuming I haven't found an easier solution or at least it hasn't been drawn to my attention I would say this is easily a p3 or p6 for an imo lvl competition

@Pathological Gotcha! Thanks

I'm the proposer. I'm glad that many people liked the problem. And thanks, terg, for posting the solution. Maybe I'll translate the original later, but you already saved me some work.

Nice problem! First we show that $f(x)\ge x$.

Click to reveal hidden text

Then we show that $f(b)>b$ cannot hold for any $b$ because in that case we would be able to produce a sequence of values of $f$ tending to $-\infty$.

Click to reveal hidden text

Only identity function work. Let $P(x,y)$ denote the assertion $f(xy+f(x)) = f(f(x)f(y))+x$. First, we give a nontrivial bound of $f$. Claim: $f(x)\geq x$ for any $x$. Proof: If $f(x)<x$, then $P\left(x,\tfrac{x-f(x)}{x}\right)$ gives $f(x) = f(\text{stuff}) + x$ which is contradiction. $\blacksquare$ From now, assume that $f(t)>t$ for some $t>0$. Call a positive real $c$ tasty if and only if there exists real $M$ which $f(x)\geq cx$ for any $x>M$. We prove two claims about tasty numbers which eventually lead us to deduce that $f$ grows faster than linear. Claim: Any $c < \tfrac{f(t)}{t}$ is tasty. Proof: Since $f(f(x)f(y))\geq f(x)f(y) \geq yf(x)$, the claim follows from $$P\left(t,\frac{x-f(t)}{t}\right)\implies f(x)\geq f(t)\cdot \frac{x-f(t)}{t} + t. \blacksquare$$ Claim: If $c$ is tasty, then any constant $c_1 < c^3$ is tasty. Proof: Indeed, if $x,y>M$, then $$f(xy+f(x)) \geq cf(x)f(y)+x \geq c^3xy+x$$By varying $y$, we see that $f(k)\geq c^3k-O(1)$ when $k>M^2+f(x)$. This gives the result. $\blacksquare$ Thus any positive real number is tasty. Therefore we can assume that $f(x)>1000x$ for any $x>M$. Now choose $\varepsilon > 0$ so that $T=\frac{f(1-\varepsilon)}{\varepsilon} > 2020M$. This is possible since $f(1-\varepsilon)\geq 1-\varepsilon$. Then consider $$P(1-\varepsilon, T)\implies f(T) \geq 1000f(1-\varepsilon)f(T) + (1-\varepsilon)$$which means $f(1-\varepsilon) < \tfrac{1}{1000}$, contradiction.

Is there proof not using ratio but proving $f(x) < x+c$ is impossible? I think it is almost impossible to do in this way.

Solution with several collaborators

: Claim: We have $f(x) \ge x$ for all $x$. Proof. Assume for contradiction $f(x) < x$ for some $x$. Then $P\left( x, \frac{x-f(x)}{x} \right)$ gives $f(x) = f(\text{blah}) + x$ or $f(\text{blah}) = f(x)-x$, contradiction. $\blacksquare$ Claim: The limit $\lim_{t \to 0} f(t)$ exists and equals zero. Proof. First, an independent lemma: we claim that $\boxed{f(x) < 1 \; \forall x < 1}$. Indeed, whenever $x < 1$, we can find $y$ such that $xy + f(x) = y$; then \[ f(y) = f(xy+f(x)) = f(f(x)f(y))+x > f(f(x)f(y)) \ge f(x)f(y) \]which implies the boxed lemma. Back to the proof of the claim. Let $0 < \varepsilon < \frac{1}{2}$ be given and choose $x = 1 - \varepsilon$. Then let $\delta$ be any number with $\delta < 1-f(x)$. Then $P(x, \delta)$ gives \[ 1 \ge f(\underbrace{f(x) + \delta \cdot x}_{<1}) = x + f(f(x)f(\delta)) > (1-\varepsilon) + f(x)f(\delta). \]Since $x > 1/2$, we have $f(x) > 1/2$. In conclusion we have prove the statement \[ f(\delta) < 2\varepsilon \qquad \forall 0 < \varepsilon < \frac{1}{2} \text{ and } \delta < 1-f(1-\varepsilon) \]So this implies the limit condition. $\blacksquare$ Fix any $x$ and consider small $y$; we have \[ x \le f(x) < xy + f(x) \le f(xy+f(x)) = x + f(f(x)f(y)). \]Since $\lim_{y \to 0} f(f(x)f(y)) = 0$ by applying the last claim twice, this implies $f(x) = x$ for all $x$.

v_Enhance wrote:

Solution with several collaborators

: How do you even coordinate this kind of thing XD

Solution - We will show that $f(x) = x$ is the only solution, firstly note that it satisfies given equation . Now let $f$ be any function satisfying the assertion which we shall call $P(x,y)$. Here goes a list of claims - Claim 1 - If $t\geq f(s)$ then $f(t) \geq s$.

Proof

Claim 2 - $f(f(s)) \geq s \hspace{0.25cm} \forall s \in \mathbb{R}_{>0}$

Proof

Claim 3 - $\frac{f(s)}{s} < f(2) \hspace{0.25cm} \forall s \in \mathbb{R}_{>0}$

Proof

Claim 4 - $f(f(x)) = x \hspace{0.25cm} \forall x \in \mathbb{R}_{>0}$ and hence it is a bijection .

Proof

Claim 5 - $f(1)=1$

Proof

Claim 6 - $f(x) = x \hspace{0.25cm} \forall x \in \mathbb{R}_{>0}$

Proof

pablock wrote: Let $\mathbb{R}_{>0}$ be the set of the positive real numbers. Find all functions $f:\mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that $$f(xy+f(x))=f(f(x)f(y))+x$$for all positive real numbers $x$ and $y$. Solution. We prove three claims. Claim 1. \(f(x)\geq x\). Proof. Assume there exists \(a\) so that \(f(a)<a\). Then, \(P\left(x,\frac{x-f(x)}{x}\right)\) gives \(f(M)=f(x)-x>0\), a contradiction. Claim 2. \(f(x)<1\) for all \(x<1\). Proof. For \(x<1\), choose \(y\) so that \(xy+f(x)=y\). Then, \[f(y)=f(xy+f(x))=f(f(x)f(y))+x>f(x)f(y)+x>f(x)f(y)\]and so we are done. Claim 3. \(\lim_{t\rightarrow0}f(t)=0\). Proof. Choose \(0<\epsilon<1\) arbitrarily small. Then \(1-\epsilon\in(0,1)\). Also choose \(\lambda\) such that \[(1-\epsilon)\lambda+f(1-\epsilon)<1\]Then, \(x=1-\epsilon\) and \(y=\lambda\) gives \[1>f(x\lambda+f(x))=f(f(x)f(\lambda))+x=f(f(x)f(\lambda))+1-\epsilon\]so \[0<f(f(x)f(\lambda))<\epsilon\]and since \(\epsilon\) is arbitrarily small, we have that \(f(f(x)f(\lambda))\) tends to \(0\). Now, \(P(x,\lambda)\) gives \[x\leq f(x)<x\lambda+f(x)\leq f(x\lambda+f(x))=f((x)f(\lambda))+x\]and since \(f(f(x)f(\lambda))\) tends to \(0\), \(f(x)=x\). Switching \(x\) and \(\lambda\) gives \(f(\lambda)=\lambda\). Therefore, we have that \(f(x)f(\lambda)=x\lambda\) (since \(f(x)=x\) for \(x<1\)) which tends to \(0\), if we choose \(\lambda\) sufficiently small. Therefore, as \(f(x)f(\lambda)\) tends to \(0\), it's image also tends to \(0\), as desired. Finally, if we fix \(x=x_0\) and choose \(y=\delta\) sufficiently small, then \[x\leq f(x)<xy+f(x)\leq f(xy+f(x))=x+f(f(x)f(y))\]Also we have \(\lim_{y\rightarrow0}f(y)=0\) so \(\lim_{y\rightarrow0}f(f(x)f(y))=0\) and so \(f(x)=x\) (since the first inequality is actually an equality)

Zelderis wrote:

Hint

I find your comment kinda interesting, I understand what u are saying, but i wonder how to present this in the paper. I mean i have done several fe, but none of them need to be proved by the definition. The only tool i have to prove the continuous is using the surjectivity and the monotonous of the function. Would you mind describing fully, like what you will write on the paper when taking the test? Appreciate a lot

hard. solved with CANNED HINTS The answer is $f(x)=x$, which clearly works Let $P(x,y)$ denote the assertion. We present a series of claims to solve this problem. Claim 1: $f(x)\geq x$. Proof: Suppose $f(x)<x$. Then $P(x, \tfrac{x-f(x)}{x})$ yields $f(x)>x$: contradiction. Claim 2: $f(x)<1$ for all $x<1$. Proof: In general for $x<1$, consider $y=\tfrac{f(x)}{1-x}$, so $xy+f(x)=y$. Then $P(x,y)$ gives $$f(y)=f(f(x)f(y))+x\geq f(x)f(y)+x \implies f(y)=\frac{x}{1-f(x)}.$$Thus $f(x)<1$, since $f(y)>0$. Claim 3: $\lim_{x \to 0} f(x)=0$. Proof: Suppose that there exists some absolute constant $C>0$ such that there exist arbitrarily small $a$ with $f(a)\geq C$. Let $\tfrac{1}{C+1}<x<1$ and pick a small $a$ as described such that $xa+f(x)<1$ as well. Then from $P(x,a)$ we have the inequality chain $$1>f(xa+f(x))=f(f(x)f(a))+x\geq f(x)C+x\geq x(C+1)>1,$$which is absurd. Thus such a $C$ does not exist, so we have $\lim_{x\to 0} f(x)=0$ as desired. Claim 4: $f(x)\leq x$. Proof: Consider $P(x,y)$ and send $y \to 0$. Then $f(x)f(y) \to 0$ as well, so $f(f(x)f(y)) \to 0$. This means that $$\lim_{y \to 0} f(xy+f(x))=\lim_{y \to 0} f(y+f(x))=x.$$On the other hand, we have $f(y+f(x))\geq y+f(x)\geq f(x)$, so if $f(x)>x$ then the above equation cannot be true. Combining claims 1 and 4 finally yields $f(x)=x$ for all $x$, as desired. $\blacksquare$

Claim 1: $f(x)\ge x$. Proof. Suppose $f(x)<x$, then $P(x,\tfrac{x-f(x)}{x}): f(x)=f(f(x)f(\tfrac{x-f(x)}{x}))+x$, contradiction. Claim 2: If $x<1$ then $f(x)<1$. Proof. From above we get $f(xy+f(x))\ge f(x)f(y)+x$. If $x<1$ we can take $y=\frac{f(x)}{1-x}$ to force $xy+f(x)=y$, which forces $f(x)<1$. Claim 3: $f$ is unbounded below. Proof. Taking $x=1-\epsilon$ and $y$ such that $xy+f(x)<1$ (possible because $f(x)<1$), we get $f(f(x)f(y))<\epsilon.$ Claim 4: There exists $f(c)=1$. Taking $x=1, f(y)\to 0^+$, we have $1<f(xy+f(x))<2$. Let $xy+f(x)=t$, so we have $1<f(t)<2$. Taking $x=f(t)-1<1$, $f(xy+f(x))+1=f(f(x)f(y))+f(t)$. It suffices to find $y$ such that $xy+f(x)=t$. Since $x<1$ and $t>1$, this is possible. Claim 5: $f(n)=n$ for all $n\in \mathbb{N}$. Proof. Taking $f(c)=1$, we have $c\le f(c)\le 1$. By Claim 2, $c$ cannot be less than 1, so $c=1$. Now $P(1,y): f(y+1)=f(f(y))+1$. By induction, $f(n)=n$. Claim 6: $f(x)\le x$. Proof. Suppose $f(x)=x+c$ for $c>0$. Then taking $y=\frac{N-x-c}{x}$ where $N\in \mathbb{N}$, we have $N=xy+x+c=f(xy+x+c)\ge f(x)f(y)+x\ge (x+c)y+x$. Taking $y$ sufficiently large leads to a contradiction. Combining claims 1 and 6, $f(x)\equiv x$, which works.

The solution splits into many steps. Step 1: Note that $f(x) \geq x$ for all $x>0$. Indeed, if $f(x)<x$ for a $x$, then $y \rightarrow (x-f(x))/x$ yields $f(x)>x$, absurd. Step 2: $f(x)<1$ for all $x<1$. Indeed, if $x<1$ then $y \rightarrow f(x)/(1-x)$ yields (using Step 1) $f(\frac{f(x)}{1-x}) \geq f(x)f(\dfrac{f(x)}{1-x})+x,$ that is $f(x)<1$ as desired. Step 3: There exists a constant $k \geq 1$ such that $f(x) \leq kx$ for all $x \leq 1$. For this step, take a $x<1$ and $y \rightarrow (f(2)-f(x))/x$. By Steps 1 and 2, $y$ is positive, and so $f(f(2)) \geq f(x) \cdot \dfrac{f(2)-f(x)}{x}+x,$ which rewrites as $\dfrac{f(x)}{x} \leq \dfrac{f(f(2))-x}{f(2)-f(x)}$ However, the RHS of the above inequality is evidently $<\dfrac{f(f(2)}{f(2)-1},$ and so $f(x)/x$ is bounded for all $x \leq 1$, as wanted. Step 4: $f(x)=x$ for all $x \leq 1$. Indeed, fix $x \leq 1$ and take $y$ to be pretty small. Then, $xy+f(x) \leq f(xy+f(x))=f(f(x)f(y))+x \leq f(x) \cdot ky \cdot k+x,$ and so $\dfrac{f(x)}{x} \leq \dfrac{1-y}{1-k^2y},$ and this fraction tends to $1$ when $y$ gets closer to $0$, and we are done. Step 5: $f$ is identity everywhere. Since $f(1)=1$, put $x=1$ to infer that $f(x+1)=f(x)+1$, and so $f(x)=f(\{ x \})+\lfloor x \rfloor=\{x \}+\lfloor x \rfloor =x,$ as wanted. It's easy to check the identity function works, and so it is the only solution.

Answer is $f \equiv x$, easy to check. Let $P(x, y)$ be the assertion above. We will prove $f(x) \geq x$ for all $x$. Assuming $f(x) < x$, take $P\left(x, \frac{x-f(x)}{x}\right)$, we obviously get $f(x) > x$. Now assuming there exists some constant $c$ such that $f(c) = c + \epsilon$, note taking $P(c, y)$ we have \[f(cy+c+\epsilon) = f((c+\epsilon)f(y)) + c\]Now taking for arbitrarily large $y$ greater than some bound, note we have $(c+\epsilon)f(y) \geq (c+\epsilon)y > cy+c+\epsilon$. Thus we can define a sequence $\{a_i\}$, such that \[a_i = \begin{cases} cy+c+\epsilon, \quad i = 1\\ (c+\epsilon)f\left(\frac{a_{i-1} - c - \epsilon}{c}\right), \quad i > 1 \end{cases} \]This gives $f(a_{i-1}) = f(a_i) + c$. Note $\{a_i\}$ is strictly increasing and $f(a_i)$ is strictly decreasing by a constant, thus a contradiction to $f: \mathbb R^+ \rightarrow \mathbb R^+$, i.e $f \equiv x$.

DS68 wrote: Answer is $f \equiv x$, easy to check. Let $P(x, y)$ be the assertion above. We will prove $f(x) \geq x$ for all $x$. Assuming $f(x) < x$, take $P\left(x, \frac{x-f(x)}{x}\right)$, we obviously get $f(x) > x$ since $\text{imf} = \mathbb R^+$. Now assuming there exists some constant $c$ such that $f(c) = c + \epsilon$, note taking $P(c, y)$ we have \[f(cy+c+\epsilon) = f((c+\epsilon)f(y)) + c\]Now taking for arbitrarily large $y$ greater than some bound, note we have $(c+\epsilon)f(y) \geq (c+\epsilon)y > cy+c+\epsilon$. Thus we can define a sequence $\{a_i\}$, such that \[a_i = \begin{cases} cy+c+\epsilon, \quad i = 1\\ (c+\epsilon)f\left(\frac{a_{i-1} - c - \epsilon}{c}\right), \quad i > 1 \end{cases} \]This gives $f(a_{i-1}) = f(a_i) + c$. Note $\{a_i\}$ is strictly increasing and $f(a_i)$ is strictly decreasing by a constant, thus a contradiction to $\text{imf} = \mathbb R^+$, i.e $f \equiv x$. Is imf the image of f? you haven't proved that f is onto

Solved with kotmhn, NTguy, rjp08 , quantam13 and Om245 Claim 1: $f(x) \geq x$ $ \forall x \in \mathbb{R}^{+}$ if $x>f(x)$ then, $P(x,x-f(x)) \implies f(x)>x$ Contradiction! Claim 2: $x<1 \implies f(x)<1$ $P(x,\frac{f(x)}{1-x}) \implies f(y) = f(f(x)f(y)) +x > f(f(x)f(y)) > f(x)f(y) \implies 1>f(x)$ Claim 3: $\lim_{x \to 1^{-}}f(x) = 1$ Trivial by Claim 1 and 2 ! Claim 4: $\lim_{x \to 0^{+}}f(x) = 0$ Let $x=1-\epsilon$, $y<1-f(1-\epsilon)$ $1 \geq xy+f(x) \implies 1 \geq f(xy+f(x)) = f(f(x)f(y)) +x$ $\implies \epsilon \geq f(f(x)f(y)) \geq f(x)f(y)$ $\implies \frac{\epsilon}{1-\epsilon} \geq f(y)$ $0<f(y)<\frac{\epsilon}{1-\epsilon}$ $\lim_{\epsilon \to 0}1-f(1-\epsilon) = 0$ (By Claim 1,2,3) y is surjective as we choose y < $1-f(1-\epsilon)$ So, $\lim_{y \to 0} f(y)=0$ Finish: $f(f(y)f(x))+x = f(xy+f(x)) \geq f(x)+xy \geq x$ $\lim_{y \to 0} f(f(x)f(y))+x=x$ $\implies \lim_{y \to 0} f(x)+xy =x \implies \lim_{y \to 0} f(x) = x$ But $f(x)$ is independent of $y $ $\implies f(x)=x$ Easy to check!

We claim $f$ is the identity, which works. First note that if $f(x)<x$ then choose $y$ such that $xy+f(x)=x$, which implies $f(x)>x$, contradiction. Thus $f(x)\ge x$ for all $x$. In particular we get $f(xy+f(x))\ge f(x)f(y)+x$. Suppose $x<1$. Then choose $y$ such that $xy+f(x)=y$ to get $f(x)<1$. Next suppose $x<n$ implies $f(x)<n$ for some $n$. If $n\le k<n+1$ then choose $x<1$ large enough such that $xy+f(x)=k$ for some $y<n$, which is possible since $f(x)>x$. Now $f(k)=f(f(x)f(y))+x<n+1$, since $f(x)f(y)<n$. Thus $x<n+1$ implies $f(x)<n+1$. By induction, this holds for all $n$. Now if $xy+f(x)<n$ then $f(x)f(y)+x\le f(xy+f(x))<n$. Setting $x<m,y=\frac{n-m}x$ gives $xy+f(x)>n-m$, so \[(f(x)f(y)+x)-(xy+f(x))=f(x)(f(y)-y)+(f(x)-x)(y-1)<m.\]The first term is positive and $y-1>n-m-1$ so $f(x)-x<\frac m{n-m-1}$. Taking $n$ arbitrarily large implies $f(x)=x$, as desired.