Does this actually work? Uh....

This was an extremely difficult but very nice FE. Solved with Ankit Bisain and Luke Robitaille.

The solution is $f(x)=x+1$ which obviously works. We now show that it is the only solution. Claim. $f$ is injective Proof. Label $$f(x+f(xy))+y=f(x)f(y)+1\hspace{20pt}(1)$$Suppose $f(a)=f(b)$ and put $(x,y)=(1,a)$ and $(1,b)$ in $(1)$ we have \begin{align*} f(1+f(a))+a&=f(1)f(a)+1\\ f(1+f(b))+b&=f(1)f(b)+1 \end{align*}Hence $a=b$ as desired. $\blacksquare$ Claim. $f$ is increasing. Proof. Suppose $f(a)<f(b)$ and $a>b$. Let $x_2=\frac{ka}{a-b}$ and $x_1=\frac{kb}{a-b}$, where $k=f(b)-f(a)$, let $y=\frac{a-b}{k}$, then $$x_1+f(x_1y)=x_2+f(x_2y)$$Hence $$f(x_1)=f(x_2)$$and so $x_1=x_2$ from injectivity, which implies $a=b$, contradiction. $\blacksquare$ Now swap $x,y$ in $(1)$ we have $$f(x+f(xy))-f(y+f(xy))=x-y\hspace{20pt}(2)$$Claim. $f(x)>1$ for all $x\in\mathbb R^+$ Proof. If $f(x)=1$, put $y=1$ in $(1)$, $$f(x+1)=f(1)$$so $x=0$, contradiction. Now from $(1)$, $$f(x)f(y)+1>y\hspace{20pt}(3)$$so $$f(y)>\frac{y-1}{f(1)}$$hence $$\lim_{y\to\infty}f(y)=\infty$$Meanwhile, $f(x)>\frac{1}{f(2)}$ by putting $y=2$ in $(3)$. If $f(y)<1$ for some $y$, define a sequence $\{x_n\}$ by $$x_{n+1}=x_n+f(x_ny)$$then $x_{n+1}\to\infty$, however, it is not hard to see $f(x_n)$ is bounded since $$f(x_{n+1})=f(x_n+f(x_ny))=f(x_n)f(y)+1-y$$hence by recursion $$f(x_{n+1})\leq f(y)^{n+1}+\frac{1-y}{1-f(y)}$$contradiction. $\blacksquare$ Claim. $$\lim_{x\to 0}f(x)=1$$Proof. We show that there exists a sequence $x_n\to 0$ such that $f(x_n)\to 1$, then the assertion clearly follows from the increasing property of the function. Let $x=y+f(1)$ in $(2)$, then $$f(y+f(1)+f(xy))-f(y+f(xy))=x-y=f(1)$$Hence there exists sufficiently large $z$ such that $f(z+f(1))-f(z)=f(1)$. Suppose on the contrary that for every positive real $n$, $f(n)\geq 1+C$ for some $C>0$, then sub. $(x,y)=(z,\frac{1}{z})$ in $(1)$, $$f(z+f(1))-f(z)=f(z)\left(f\left(\frac{1}{z}\right)-1\right)+1-\frac{1}{z}>cf\left(\frac{1}{z}\right)+1-\frac{1}{z}\hspace{20pt}(4)$$contradiction. $\blacksquare$ Now put $x=y$ in $(1)$ and let $x,y\to 0$, then $$\lim_{x\to 1}f(x)=2$$Therefore, letting $x\to 0$ in $(1)$ we have $$2+y=f(y)+1$$hence $f(y)=y+1$ as desired.

After injectivity and increasing of function $f$ it can be done as follows Lemma: From monotonous of function $f$ it follows that, for any $t\ge 0$ there exists $\lim_{x\to t^{+}}f(x) \ge 0.$ Let $A = \lim_{x\to 0^{+}}f(x)$ and $B=\lim_{x\to A^{+}}f(x).$ Fix some $c\in\mathbb{R^{+}}.$ $$B+c = \lim_{x\to 0^{+}}(f(x+f(xc))+c=\lim_{x\to 0^{+}}(f(x)f(c)+1) = Af(c)+1$$ Therefore $f$ is linear. Further part of solution is quite easy.

nukelauncher wrote: Solved with Ankit Bisain and Luke Robitaille.

Easy for A8

. Can someone give me a hint on what to do next?

/bump Anyone has a hint?

Show it's strictly increasing, and then consider the lim $p$ of $f(x)$ when $x$ approaches $0$, and the lim $q$ of $f(x)$, when $x$ approaches $p$. Use this to prove linearity (consider what happens when you let $x$ approach $0$ in the FE)

VicKmath7 wrote: Show it's strictly increasing, and then consider the lim $p$ of $f(x)$ when $x$ approaches $0$, and the lim $q$ of $f(x)$, when $x$ approaches $p$. Use this to prove linearity (consider what happens when you let $x$ approach $0$ in the FE) Thanks, I will try to do this.

Hardest F.E. i solved on my life. Denote $P(x,y)$ the assertion of the given F.E. Claim 1: $f$ is injective Proof: Let $a,b$ be positive reals such that $f(a)=f(b)$. Then by $P(1,a)-P(1,b)$ $$f(1+f(a))-f(1+f(b))+a-b=f(1)f(a)-f(1)f(b)+1-1 \implies a=b \implies f \; \text{injective}$$Claim 2: $f$ is strictly increasing Proof: Assume that there exists $m,n$ such that $m>n$ and $f(n)>f(m)$ then: $P \left(\frac{nf(n)-nf(m)}{m-n}, \frac{m-n}{f(n)-f(m)} \right)-P \left(\frac{mf(n)-mf(m)}{m-n}, \frac{m-n}{f(n)-f(m)} \right)$ $$f \left( \frac{nf(n)-nf(m)}{m-n} \right)=f \left( \frac{mf(n)-mf(m)}{m-n} \right) \implies m=n \implies \text{contradiction!!}$$We used Claim 1 to show that $f$ is strictly increasing becuase its increasing+injective. Claim 3: $f$ is lineal. Proof: Since $f$ is strictly increasing it makes sence the replace $f(x)=\lim_{x \to z^+} h(z)$ where $z \ge 0$. Thus now replacing $x \to 0^+$ on the original F.E. $$h(h(0))+y=h(0)f(y)+1 \implies f \; \text{lineal}$$Claim 4: $f(x)=x+1$ for every positive real $x$ Proof: We have that $f(x)=ux+v$ where $u,v$ are positive real numbers, then replacing: $$ux+uv+v+y=uvx+uvy+v^2+1 \overset{x=y=1}{\implies} (u+v)=(u+v)v \implies v=1 \implies f(x)=ux+1$$$$u+y=uy+1 \overset{y=2}{\implies} u=1 \implies f(x)=x+1$$Thus we are done

mathaddiction wrote: however, it is not hard to see $f(x_n)$ is bounded since $$f(x_{n+1})=f(x_n+f(x_ny))=f(x_n)f(y)+1-y$$hence by recursion $$f(x_{n+1})\leq f(y)^{n+1}+\frac{1-y}{1-f(y)}$$ . Sorry but I don't understand this step. Can anybody help to clarify?

Hopeooooo wrote: Let $R+$ be the set of positive real numbers. Determine all functions $f:R+$ $\rightarrow$ $R+$ such that for all positive real numbers $x$ and $y$ $f(x+f(xy))+y=f(x)f(y)+1$ $(Ukraine)$ Oh my god. This is one of the hardest fes I have ever done in my life. Not only is it hard, but it is also the type of fe which is trivialized when the domain adds a \(0\). Since I love this problem, I will try to give some motivations. Let \(P(x,y)\) denote the assertion of the given functional equation. \(P(1,y)\) gives us that \(f\) is injective. Now, we know that if \(f\) is strictly increasing, then we can let \(f(x)=\lim_{x\rightarrow0}g(x)\), because \(f\) is increasing and \(g(0)\neq0\). So we have \[g(g(0))+y=g(0)f(y)+1\]implying that \(f\) is linear. Now we should show that \(f\) is strictly increasing. One idea we could try to use is that if \(m>n\) and \(f(m)<f(n)\), we should try to get \(m=n\) through some trick. Now, looking at the functional equation, substituting \(x\) or \(y\) as \(m\) will definitely not yield anything useful. So we now move to the next possibility, that is letting \(xy\) to be \(m\) or \(n\). We take the assertions \(P\left(x,\frac{m}{x}\right)\) and \(P\left(y,\frac{n}{y}\right)\). These assertions yield us \[f(x+f(m))+\frac{m}{x}=f(x)f\left(\frac{m}{x}\right)+1\]and \[f(y+f(n))+\frac{n}{y}=f(y)f\left(\frac{n}{y}\right)+1\]Now, if we let \(\frac{m}{x}=\frac{n}{y}\), then comparing the equations we got, the \(\frac{m}{x}\) and \(\frac{n}{y}\) terms cancel off and the \(f\left(\frac{m}{x}\right)\) term is common. Now, if we also assume that \(x+f(m)=y+f(n)\), we get that \(f(x)=f(y)\) and so \(x=y\) by injectivity! Now we should solve the system of equations \[\begin{cases} \frac{m}{x}=\frac{n}{y} \\ x+f(m)=y+f(n) \end{cases}\]to get \((x,y)=\left(\frac{m(f(n)-f(m))}{m-n},\frac{n(f(n)-f(m))}{m-n}\right)\) and \(x,y\) are both positive too, so we can put them in the assertion. So, we have that \(f\left(\frac{n(f(n)-f(m))}{m-n}\right)=f\left(\frac{m(f(n)-f(m))}{m-n}\right)\) so \(m=n\), a contradiction, thus \(f\) is strictly increasing. Now that \(f\) is strictly increasing, from the first paragraphs, we see that \(f\) is linear. Checking the original assertion yields that \(f(x)=x+1\) for all positive reals \(x\), and we are done.

PS: I edited the solution thanks to Luke Robitaille (I missed a crucial detail)

mathaddiction wrote: Claim. $f(x)>1$ for all $x\in\mathbb R^+$ Proof. If $f(x)=1$, put $y=1$ in $(1)$, $$f(x+1)=f(1)$$so $x=0$, contradiction. Um what if \(f(x)<1\)? You have not discussed that case

http://www.imo-official.org/problems.aspx

geometryzeus wrote: http://www.imo-official.org/problems.aspx Uhh what do you want to convey?

Let $P(x,y)$ denote the condition in the question. Solution. Claim 1: $f$ is injective. Proof: Suppose $f(a)=f(b)$. Comparing $P(1,b)$ with $P(1,a)$ forces $a=b$, as needed. Claim 2: $f$ is increasing. Proof: Fix $y$, move $x$ We know if $x_1\ne x_2$ Then $f(x_1+f(x_1y)) \ne f(x_2+f(x_2y))$ $x_1+f(x_1y)\ne x_2+f(x_2y)$ Let $t_1=x_1y, t_2=x_2y$ then this becomes $\frac{t_1-t_2}{y} = f(t_2)-f(t_1)$. If $f(a)>f(b)$ but $a<b$ we can have $t_2=a, t_1=b, y=\frac{b-a}{f(a)-f(b)}$. With this in mind, we know $f(x+f(xy))>f(x)$ Therefore, $y<f(x)(f(y)-1)+1$ $f(y)>1+\frac{y-1}{m}$ where $m=\lim\limits_{x\to 0} f(x)$. Claim 3: $m= 1$. Suppose $m>1$. Then let $a=f(1)$. $P(x,x^{-1})$ yields $f(x+a)=1-\frac 1x + f(x)f(x^{-1}) > 1-\frac 1x +m^2$. Denote this assertion $Q(x)$ Consider a chain $x\rightarrow x+a\rightarrow x+2a\rightarrow \cdots$. By induction, we can show if $x>1$, then $f(x+ta)>m^t$. This implies $f$ is greater than exponential growth. Finish 1: Set $x=y$ to get $f(x)>\frac 12 \sqrt{f(x+f(x^2)} = \frac 12 \sqrt{c^{x+f(x^2)}} > \frac 12 c^{\frac 12(x+c^{x^2})}$ for some $c>1$ and $x$ large enough. Now, we repeat this argument for $x=y$ to get arbitarily large bounds on size of $f$ ). In particular, if we fix $f(t)=a$ for any $t$ we can get a contradiction after a couple of "recursions". We can also increase $m$ by setting $x\rightarrow 0$. Finish 2. Consider another chain $x_0=x, x_n=x_{n-1}+f(x_{n-1})$. Note $P(x_n,1)$ yields $f(x_n)=f(x_{n-1})f(1)$ so $f(x_n)=f(x)f(1)^n$. However, $x_n$ is also exponential, so $f$ is near linear at $x_n$, contradiction. Suppose $m<1$ then $f(y)>1+\frac{y-1}{m}$. Using $f(x+f(xy))-(x+f(xy))=f(y+f(xy))-(y+f(xy))$ we can get similar arbitrarily large bounds for f. Therefore, $m=1$. Now, setting $x\to 0$, $P(x,x)$ gives $\lim_{x\to 0} f(x+1)=2$ Therefore, $P(x\rightarrow 0,y)$ gives $f(1)+y=1f(y)+1$, so $f(y)\equiv y+1$.

Claim: $f$ is injective. Proof: $P(1,x): f(1+f(x))+x=f(1)f(x)+1$. If $f(a)=f(b)$, then do $P(1,a)-P(1,b)$ to get $a=b$. $\blacksquare$ Claim: $f$ is strictly increasing Proof: Suppose there are $a$ and $b$ with $a>b$ and $f(b)>f(a)$ (since $f$ is injective, showing that this is not possible proves our claim). Note that $c=\frac{a-b}{f(b)-f(a)}$ is positive. $P\left(\frac{a}{c},c\right): f\left(\frac{a}{c}+f(a)\right)+c=f\left(\frac{a}{c}\right)f(c)+1$. We have \[\frac{a}{c}+f(a)=\frac{af(b)-af(a)+af(a)-bf(a)}{a-b}=\frac{af(b)-bf(a)}{a-b}.\]We also have \[\frac{b}{c}+f(b)=\frac{bf(b)-bf(a)+af(b)-bf(b)}{a-b}=\frac{af(b)-bf(a)}{a-b}=\frac{a}{c}+f(a)\]So doing $P\left(\frac{b}{c},c\right)$ gives $f\left(\frac{a}{c}\right)=f\left(\frac{b}{c}\right)$, so $a=b$. $\blacksquare$ Claim: $f$ is linear Let $p$ be the limit when $f(x)$ approaches $0$ and $q$ be the limit when $f(x)$ approaches $p$ from above. We will show that $q+y=pf(y)+1$, which implies $f$ is linear. Set $x$ arbitrarily close to $0$. Then we get $f(x+p)+y=pf(y)+1\implies q+y=pf(y)+1$. $\blacksquare$ \begin{align*} P(1,x): f(ax+b+1)+x=(ax+b)(a+b)+1 \\ \implies a^2x+ab+a+b+x=a^2x+ab+b^2+axb+1 \\ \implies a+b+x=b^2+axb+1 \\\implies ab=1\\ \text{and } a+b=b^2+1 \\ \end{align*}Now we get $a+\frac{1}{a}=\frac{1}{a^2}+1$. Multiplying by $a^2$ gives $a^3-a^2+a-1=0$. So $(a-1)(a^2+1)=0$. Since $a$ is real, $a=b=1$ and the only solution is $\boxed{f(x)=x+1}$, which works.

The hardest part was proving $f(x)=ax+b,$ and I think this defines the difficulty of the whole problem, well at least in my case. Nevertheless thank you Ukraine for the problem.

ZETA_in_olympiad wrote: Claim 2: $f$ is a non-decreasing monotonic function. Proof. From injectivity it follows, $$\forall a,b \in \mathbb{R^+}: a\neq b\implies a+f(ay)\neq b+f(by).$$ @ZETA_in_olympiad be careful! Injectivity and the fact that $a\neq b$ do not necessarily imply $a+f(ay)\neq b+f(by).$ What if $a=1,b=2$ and $f(ay)=2,f(by)=1$ for example? Generally speaking, if $x\neq y$ and $m\neq n$ we do not necessarily have $x+m\neq y+n.$ ZETA_in_olympiad wrote: $n+y=mf(y)+1\implies f(y)=\frac{n+y-1}{m} ~~\forall m\neq 0.$ As a side note, you should clearly state that due to $f$ being non-decreasing, we cannot have $m=n=\infty,$ so we also cannot have $m=\infty$ or $n=\infty.$ You also have to make sure that $m\neq 0$ and only then can you say that $f(y)=(n+y-1)/m.$ I am unsure what you meant by $\forall m\neq 0$, as $m$ is a constant.

oVlad wrote: I am unsure what you meant by $\forall m\neq 0$, as $m$ is a constant. $m=0$ means $n+y=1$ which is false. I realize "for all" is absurd here.

oVlad wrote: ZETA_in_olympiad wrote: Claim 2: $f$ is a non-decreasing monotonic function. Proof. From injectivity it follows, $$\forall a,b \in \mathbb{R^+}: a\neq b\implies a+f(ay)\neq b+f(by).$$ @ZETA_in_olympiad be careful! Injectivity and the fact that $a\neq b$ do not necessarily imply $a+f(ay)\neq b+f(by).$ What if $a=1,b=2$ and $f(ay)=2,f(by)=1$ for example? Generally speaking, if $x\neq y$ and $m\neq n$ we do not necessarily have $x+m\neq y+n.$ Fixed it (?) Thanks for pointing. I don't know if there are more loopholes.

megarnie wrote: Claim: $f$ is linear Let $p$ be the limit when $f(x)$ approaches $0$ and $q$ be the limit when $f(x)$ approaches $p$ from above. We will show that $q+y=pf(y)+1$, which implies $f$ is linear. Set $x$ arbitrarily close to $0$. Then we get $f(x+p)+y=pf(y)+1\implies q+y=pf(y)+1$. $\blacksquare$ \begin{align*} P(1,x): f(ax+b+1)+x=(ax+b)(a+b)+1 \\ \implies a^2x+ab+a+b+x=a^2x+ab+b^2+axb+1 \\ \implies a+b+x=b^2+axb+1 \\\implies ab=1\\ \text{and } a+b=b^2+1 \\ \end{align*}Now we get $a+\frac{1}{a}=\frac{1}{a^2}+1$. Multiplying by $a^2$ gives $a^3-a^2+a-1=0$. So $(a-1)(a^2+1)=0$. Since $a$ is real, $a=b=1$ and the only solution is $\boxed{f(x)=x+1}$, which works. Hello, Sorry for my question but why do you know that there exists that two limits? There is no hypothesis for the continuity of $f$

nguyenvuthanhha wrote: megarnie wrote: Claim: $f$ is linear Let $p$ be the limit when $f(x)$ approaches $0$ and $q$ be the limit when $f(x)$ approaches $p$ from above. We will show that $q+y=pf(y)+1$, which implies $f$ is linear. Set $x$ arbitrarily close to $0$. Then we get $f(x+p)+y=pf(y)+1\implies q+y=pf(y)+1$. $\blacksquare$ \begin{align*} P(1,x): f(ax+b+1)+x=(ax+b)(a+b)+1 \\ \implies a^2x+ab+a+b+x=a^2x+ab+b^2+axb+1 \\ \implies a+b+x=b^2+axb+1 \\\implies ab=1\\ \text{and } a+b=b^2+1 \\ \end{align*}Now we get $a+\frac{1}{a}=\frac{1}{a^2}+1$. Multiplying by $a^2$ gives $a^3-a^2+a-1=0$. So $(a-1)(a^2+1)=0$. Since $a$ is real, $a=b=1$ and the only solution is $\boxed{f(x)=x+1}$, which works. Hello, Sorry for my question but why do you know that there exists that two limits? There is no hypothesis for the continuity of $f$ f is strictly increasing

why is f strictly increasing?

DottedCaculator wrote: why is f strictly increasing? because i proved f is strictly increasing

DottedCaculator wrote: why is f strictly increasing? User nguyenvuthanhha did cut the $f$ is strictly increasing part in his quote.

This seems diferent to the above, I hope it is correct. I really enjoyed this FE! I claim the only such function is $f(x)=x+1$ for all $x\in \mathbb{R}^+$, which clearly works. I now prove that this is the only one. My proof procedes in four steps. First, I find $f(1)$ and solve the functional equation over the positive rationals. Then, I prove injectivity and strict monotonicity to finally extend my result to $\mathbb{R}^+$ and conclude. Step 1: We find $f(1)=2$. For now, let $a:=f(1)$. I claim the following Claim: In fact, it holds that $$k:=\inf_{y\in \mathbb{R}^+} \left \{ \frac{f(y)}{y+1} \right\}=a-1$$Pf: First I claim that $k\leq a-1$. If not, then by definition of $a, k$ we find \begin{align} P(x, 1): af(x)=f(x+f(x))\geq k(x+f(x)+1)\implies \frac{f(x)}{x+1}\geq \frac{k}{a-k} \text{ for all } x\in \mathbb{R}^+ \label{eq:cooleq} \end{align}but by our assumption $a-k<1$ and we must have $$\frac{k}{a-k}>k$$which contradicts the fact that $k$ is the infimum. Now I claim $k\geq a-1$, which proves that $k=a-1$. Consider the sequence $\{k_i\}_{i=1}^{\infty}$ defined by $$k_1=k \text{ and } k_{n+1}=\frac{k_n}{a-k_n} \text{ for all } n\geq 0, $$which has limit $a-1$. Notice that Equation (1) implies that $$\frac{f(x)}{x+1}\geq k_n \text{ for every } x\in \mathbb{R}^+ \text{ and } n\in \mathbb{N}$$and thus we must also have $$\frac{f(x)}{x+1}\geq \lim_{n\to\infty} k_n=a-1.$$In particular, by definition of infimum, we must have $k\geq a-1$, as desired. $\blacksquare$ Now see that $$P(1, x): af(x)+1=x+f(1+f(x))\geq x+(a-1)(1+f(x)+1)\implies \frac{f(x)}{x+1}\geq 1+\frac{2(a-2)}{x+1}$$Since we have seen that $k=a-1$, applying this result to the above yields $$1+\frac{2(a-2)}{x+1}\leq a-1 \text{ for every } x\in \mathbb{R}^+\implies (a-2)(x-1)\geq 0 \text{ for every } x\in \mathbb{R}^+$$which is a contradiction unless we have exactly $a=2$. Step 2: We have $$f(q)=q+1 \text{ for every } q\in \mathbb{Q}^+$$ The proof is a straightforward induction starting with the base case $f(1)=2$, proven in Step~1. Step 3: $f$ is strictly monotone increasing (and injective). Injectivity follows from $P(1, x)$. Since $f$ is injective, we must have $$x_1+f(x_1y)=x_2+f(x_2y)\implies x_1=x_2$$for a fixed $y\in \mathbb{R}^+$, by looking at the LHS of $P(x_1, y)$ and $P(x_2, y)$. In particular, this means that we can never have $$\frac{1}{y}=\frac{f(x_1y)-f(x_2y)}{-(x_1y-x_2y)} \text{ when } x_1y\neq x_2y$$implying that $f(a)-f(b)$ and $a-b$ have the same sign, as desired. Step 4: $f(x)=x+1$ for all $x\in \mathbb{R}^+$ We will prove that given that we know values at all rational points and monotonicity, we can conclude that $f(x)\equiv x+1$. Suppose that for some $x\in \mathbb{R}^+$ we have $f(x)\neq x+1$. Then we can find some rational $q$ such that $$x+1 < q=f(q-1)<f(x)$$But since $f$ is monotone increasing we must have $$q-1<x,$$a contradiction to $x+1<q$. Hence we have proven that $f(x)=x+1$ is the only function that satisfies the problem statement over $\mathbb{R}^+$ $\square$

Let $P(x,y)$ denote the assertion. We have $f(1,y)$ implies $f(1+f(y))+y=f(1)f(y)+1$, so $f$ is injective. Next, consider $P(\tfrac{x}{a},x)$ which gives \[f\left(\frac{x}{a}+f(x)\right)+a=f\left(\frac{x}{a}\right)+1\]Thus, if $\tfrac{x}{a}+f(x)=\tfrac{y}{a}+f(y)$ then $x=y$. Suppose $x>y$ then $\tfrac{x}{a}+f(x)\neq \tfrac{y}{a}+f(y)$ for any $y$. If $f(y)<f(x)$ then $a$ must exist to make that equation come true, so $f(y)>f(x)$, implying that $f$ is increasing. Let $x^+$ be a number that approaches $x$ from the positive side. In particular, $f(x^+)=\lim_{t\to x^+}(f(t))$. Note that $0^++c=c^+$, so $P(0^+,y)$ gives $f(f(0^+)^+)+y=f(0^+)f(y)+1$ which implies that $f$ is linear. Let $f(x)=ax+b$ where $a>0, b\ge 0$ then $ax+a^2xy+ab+b+y=a^2xy+abx+aby+b^2+1$ for all positive $x$ and $y$ which clearly implies $a=b=1$. From our derivation of $f$ is is clear that $f(x)=x+1$ works.

Someone please explain this for me, i agree that fix y and taking x to 0+ will exist limf(x)=p x->0+ and lim f(x+p)=q x->0+ But the step when x->0+ ,f(xy) -> p, $limf(x+f(xy)) = lim(x+p)$ would require f is continuous, right?(the step written in latex)

One of the hardest FE I've solved . The only function that satisfies is $f(x)=x+1 \ \forall \ x \in \mathbb{R^+}$. Clearly this function works, now we'll prove that only this function satisfies. Let $P(x,y)$ denote the assertion. All variables mentioned should be in $\mathbb{R^+}$ unless specified. Claim 1. $f$ is injective Proof. If $f(a)=f(b)$, then $P(1,a)$ and $P(1,b)$ implies that $a=b$. As desired. Claim 2. $f(1)=2$ Proof. Let $f(1)=c$. First, from $P(1,y)$ we have $$f(1+f(y))=cf(y)+1-y \ \ (1)$$. While from $P(x,1)$ we have $$(x+f(x))=cf(x) \ \ (2)$$. Plug $x=1$ to $(2)$ and get $f(1+c)=c^2$ plug $x=1+c$ to $(1)$ again to get $f(1+c+c^2)=c^3$, do this one more time and get $$f(1+c+c^2+c^3)=c^4 \Rightarrow f((1+c)(1+c^2))=c^4$$.Next, plug $y=1+c$ to $(1)$ and get $f(1+c^2)=c^3-c$ and plug $y=1+c+c^2$ to $(2)$ and get $f(1+c^3)=c^4-c^2-c$. Now from $P(1+f(1),1+f(1)^2)$, we have $$f(1+c+c^4)=c(c^4-c^2-c)$$On the other hand, if we plug $x=1+c^3$ to $(2)$ we will get $$f(c^4+c^3-c^2-c+1)=c(c^4-c^2-c)$$. By injectivity we get $$c^3-c^2-c=c \Rightarrow c(c+1)(c-2)=0$$. Hence $f(1)=c=2$. As desired. Claim 3. $f$ is strictly increasing Proof. For $x<y$, assume $f(x)>f(y)$. Note that $P\left( \frac{x(f(x)-f(y))}{y-x},\frac{y-x}{f(y)-f(x)} \right)$ shows $$f\left( \frac{yf(x)-xf(y)}{y-x} \right)+ \frac{y-x}{f(y)-f(x)}=f\left( \frac{x(f(x)-f(y))}{y-x}\right)f\left( \frac{y-x}{f(y)-f(x)} \right)+1$$. Where the LHS is symmetric in $x,y$ so swapping them and using injectivity gives $x=y$. A contradiction. Hence we must have $f(x)<f(y)$ since $f(x) \neq f(y)$. Claim 4. $f(x)>1$ Proof. We can use $(2)$ and $f(1)=2$ to inductively prove $f(2^n-1)=2^n$for all $n \in \mathbb{N}$. Then from $P(x,2^n-1)$ we have $$f(x)2^n+1>2^n-1 \Rightarrow f(x)>\frac{2^n-2}{2^n}$$. Setting $n$ to $\infty$ gives $f(x)\ge 1$. If $f(x)=1$ for some $x$ then plug that $x$ to $(2)$ and get $f(1+x)=2=f(1)$. Contradiction. As desired. Claim 5. $\lim_{x \to 0} f(x)=1$ Proof. From $(1)$ we know $\lim_{x \to \infty} f(x)=\infty$. And also since $f(x)>1$,$\lim_{x \to 0} f(x)\ge 1$. Assume $\lim_{x \to 0} f(x) > 1$, then $f(x)\ge 1+c$ for some constant $c \in \mathbb{R^+}$ and all $x \in \mathbb{R^+}$. From $P\left( x,\frac{1}{x} \right)$, we have $$f(x+2)+\frac{1}{x}=f(x)f\left( \frac{1}{x} \right)+1 \Rightarrow f(x+2)-f(x) \ge cf(x)+1-\frac{1}{x} $$. On the other hand from $P(x+2,x)-P(x,x+2)$ we have $$f(x+2+f(x^2+2x))-f(x+f(x^2+2x))=2$$. So $f(z+2)-f(z)=2$ for some arbitrarily large $z$. Set $x=z$ to the inequality and set $\lim_{z \to \infty} z$ to get a contradiction. Note that $(2)$ implies $\lim_{x \to 1} f(x)=2$, so $\lim_{x \to 0}f(x+f(xy))=2$, lastly set $\lim_{x \to 0}$ at $P(x,y)$ and get $$f(1)+y=f(y)+1 \Rightarrow f(y)=y+1$$. And we are finished.