It is from the short-list 2005 (i think A2) and it's offisial solution is very beautiful

You can also see this problem similar it http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=f%28x%29f%28y%29%3Df%28x%2Byf%28x%29%29&t=14110[/list]

I think the only solution is the constant function $f\equiv 2$. Suppose we can find $x$ such that $f(x)<1$. Then $y=\frac x{1-f(x)}>0$, and $x+yf(x)=y$. Plug these values of $x,y$ into the equation to get $f(x)f(y)=2f(y)\Rightarrow f(x)=2$, contradicting $f(x)<1$. We thus find $f(x)\ge 1,\ \forall x>0\ (*)$. Now assume we can find $y$ such that $f(y)<2$. Choose $x_0>0$ arbitrarily, and put $x_{n+1}=x_n+yf(x_n),\ \forall n\ge 0$. We have $f(x_n)=\left(\frac{f(y)}2\right)^nf(x)\to 0$ as $n\to\infty$, but this contradicts $(*)$ applied to $x_n$. This means that $f(x)\ge 2,\ \forall x>0\ (**)$. $(**)$ applied to $y$ instead of $x$, together with the initial equation, show that $f(x)\le f(x+yf(x)),\ \forall x,y>0$, i.e. $f$ is non-decreasing. If $x_1<x_2$ and $f(x_1)=f(x_2)$ (i.e. if $f$ is not one-to-one), then it follows from the equation that $\forall t>0,\ f(x_1+t)=f(x_2+t)$, and together with the fact that $f$ is non-decreasing, this implies that $f$ is eventually constant. This constant can only be $2$, so $f\equiv 2$. Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$, we get $x+yf(x)=y+xf(y),\ \forall x,y>0$. This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$, but this contradicts $(**)$.

#20002 pwns wrote:

Nope, $0$ is not in the domain ($\mathbb R^+=\{x\in\mathbb R|x>0\}$)

grobber wrote: Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$, we get $x+yf(x)=y+xf(y),\ \forall x,y>0$. This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$, but this contradicts $(**)$. Can you explain this part

abdurashidjon wrote: grobber wrote: Now assume that $f$ is not constant. From the paragraph above we see that $f$ must be one-to-one. Since $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$, we get $x+yf(x)=y+xf(y),\ \forall x,y>0$. This means that for some $a>0,\ f(x)=ax+1,\ \forall x>0$, but this contradicts $(**)$. Can you explain this part @abdurashidjon I think he replaced $x$ by $y$ and $y$ by $x$ and got $2f(x+yf(x))=f(x)f(y)=2f(y+xf(y))$, and from property one to one he got $x+yf(x)=y+xf(y)$ $\Longrightarrow \frac{f(x)-1}{x}=\frac{f(y)-1}{y}=constant=a$ So $f(x)=ax+1$,contradiction.

@Cefer thanks for answer

Let $P(x,y)$ be the assertion that $f(x)f(y) = 2f(x+yf(x))$. $P(x+zf(x), y)$ yields $f(x+zf(x))f(y) = 2f(x+zf(x) + yf(x+zf(x))$. $f(x+zf(x)) = \frac{f(x)f(z)}{2}$, so $f(x)f(y)f(z) = 4f(x+zf(x) + \frac{yf(x)f(z)}{2})$. Similarly, $P(x, z+yf(z))$ yields $f(x)f(y)f(z) = 4f(x+(z+yf(z))f(x)) = 4f(x + zf(x) + yf(x)f(z))$. Hence, $f(x+zf(x) + \frac{yf(x)f(z)}{2}) = f(x+zf(x) + yf(x)f(z))$ for all positive $x,y,z$. Let $a$ and $b$ be any positive reals such that $2a > b > a$. Let $x = a - \frac{b}{2}$, $z = \frac{2a-b - x}{f(x)}$, and $y = \frac{2(b-a)}{f(x)f(z)}$. Then $x + zf(x) + \frac{yf(x)f(z)}{2} = 2a-b + (b-a) = a$, and $x + zf(x) + yf(x)f(z) = 2a-b + 2(b-a) = b$, so $f(a) = f(b)$. It follows that $f$ is constant over all intervals of the form $(c, 2c)$. For all integers $k$, let $S_k = ((3/2)^k, 2 (3/2)^k)$. From the above, $f$ is constant over each $S_k$. Since $2(3/2)^k > (3/2)^{k+1}$, $S_{k+1} \cap S_k \neq \emptyset$, so the constant value of $f$ over any $S_k$ is equal to that of $S_{k+1}$. Since $\bigcup_{k=-\infty}^{\infty} S_k = \mathbb{R}^+$, $f$ must be constant over $\mathbb{R}^+$. It follows that $f(x) = 2$ for all $x$.

Here is a solution I came up with a long time ago: Note that from the given, we have: \[f\left(x\right)=\frac2{f\left(y\right)}f\left(x+yf\left(x\right)\right)\] Thus we have: \begin{eqnarray*}2f\left(x\right)f\left(y+2z\right)&=&f\left(x\right)f\left(y\right)f\left(\frac{2z}{f\left(y\right)}\right)\\ &=&2f\left(x+yf\left(x\right)\right)f\left(\frac{2z}{f\left(y\right)}\right)\\ &=&4f\left(x+yf\left(x\right)+\frac{2z}{f\left(y\right)}f\left(x+yf\left(x\right)\right)\right)\\ &=&4f\left(x+yf\left(x\right)+zf\left(x\right)\right)\\ &=&2f\left(x\right)f\left(y+z\right) \end{eqnarray*} Thus dividing both sides by $f\left(x\right)$ gives $f\left(y+z\right)=f\left(y+2z\right)$ for all positive reals $y,z$. Now repeatedly replacing $y$ with $y+z$ gives $f\left(y+z\right)=f\left(y+nz\right)$ for all reals $y,z$ and positive integers $n$. Now suppose $a,b$ are arbitrary positive real numbers with $a>b$. Let $n$ be an integer greater than $a/b$, and let \[y=\frac{nb-a}{n-1}, z=\frac{a-b}{n-1}\] Thus $y+nz=a$ and $y+z=b$, so we have $f\left(a\right)=f\left(b\right)$. However, since $a,b$ were arbitrary, $f\left(x\right)\equiv c$ for some $c$, so plugging this back into the given equation gives $f\left(x\right)\equiv 2$ as the only solution. $\blacksquare$

Mewto55555 wrote: Thus, $f$ is periodic. Let it have period $p \ge 0$ (if $p=0$, $f$ is constant) [...] $\implies 2f(\frac{p}{2})=2f(p) \implies f(\frac{p}{2})=f(p) \implies p|\frac{p}{2}$, which is impossible. This is only a contradiction if there exists a minimal period $p$, which is not necessarily true. (You don't rule out the case in which $f$ takes one value over the rationals and another over the irrationals, in which case $p$ could be any rational number.)

Whoops, but that's easily fixed if we replace that last $\frac{p}{2}$ with an $x$, meaning $p$ divides all real $x$, impossible.

I don't see how you're getting that $p$ divides all real $x$. Your solution doesn't address the case I mentioned in my last post. Edit: Actually, I missed an even more glaring mistake: Mewto55555 wrote: Thus, $f(x+yf(x))=f(x+yf(x)+pf(x)) \implies p|pf(x)$, so $f(x)$ is a positive integer for all $x$. You're going to need some more algebraic manipulation with the original equation to get more information.

We can also think of the generalization : f(x)f(y)=af(x+yf(x)) where a>1, which nevertheless can be solved in the same manner. The case a=1 needs an additional assumption to be solved (look here : http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=f%28x%29f%28y%29%3Df%28x%2Byf%28x%29%29&t=14110 ). And I didn't look for the cases 0<a<1.

I solved it by looking at the value of f(1) and manipulating. Replacing x by x+f(x) was the key step

Pascal96 wrote: I solved it by looking at the value of f(1) and manipulating. Replacing x by x+f(x) was the key step Please post the full solution.

The solution I spoke of earlier was wrong. Sorry, I should have deleted my post. Anyway, I tried the problem again much later, and I made some partial progress but then got stuck. Basically I assumed to the contrary that f is not constant and then obtained that f is periodic. I don't remember the details now but I ran into a small problem as follows: let T be the period. I proved that f is periodic with period T/2^n for any n, so the period is infinitely small. I thought that this was it and the problem was done, but then someone said that if T was irrational, no matter how large n is the limit will be irrational and hence non zero.

Thanks to applepi2000 for some corrections

Nima Ahmadi Pour wrote: We denote by $\mathbb{R}^+$ the set of all positive real numbers. Find all functions $f: \mathbb R^ + \rightarrow\mathbb R^ +$ which have the property: \[f(x)f(y)=2f(x+yf(x))\]for all positive real numbers $x$ and $y$. Proposed by Nikolai Nikolov, Bulgaria

Nvm, I didn't read the first part.

@above We also have that f is periodic, hence it either must decrease again after increasing (contradiction) or it will be constant from then on. It's easily checked that c^2=2c where f(x)=c

Solution w/o proving non-decreasing: Let $P(x,y)$ denote the given assertion. Comparing $P(x,y)$ and $P(y,x)$ gives that $f(x+yf(x))=f(y+xf(y))$. If $f$ is injective, then $x+yf(x)=y+f(x)$. Subbing $y=1$ gives that $f$ is linear, and it is easy to check that no solutions work. So $f$ is non-injective, and there exist $a<b$ with $f(a)=f(b)=k$ Comparing $P(a,y)$ and $P(b,y)$ gives $f(a+yk)=f(b+yk) \implies f(y)=f(y+b-a)$ for $y>a$. Let $b-a=T$. Let $n$ be a positive integer. For $x>a$, $P\left(x,\frac{nT}{f(x)}\right)\implies f\left(\frac{nT}{f(x)}\right)=2$ So there exist sufficiently large $\alpha$ with $f(\alpha)=2$. Taking $x=\alpha$ gives $f\left(\frac{nT}2\right)=2$ for any $n$. $P\left(\frac{nT}2,y\right) \implies f(y)=f\left(2y+\frac{nT}2\right)$ Letting $n>4$, $f(y)=f\left(2y+\frac{nT}2\right)=f\left(2y+\frac{(n-4)T}2+2T\right)=f(y+T)$, so $f$ is periodic for all $y$. Then setting $n=2$ gives that $f(y)=f(2y+T)=f(2y)$. For any $x$, let $m$ be a positive integer such that $2^m-f(x)>0$. Then $P\left(x,\frac x{2^m-f(x)}\right) \implies f(x)\equiv 2$, which works.

Solved with john0512. The answer is $f\equiv 2$ which clearly works. Claim: $f(x)\ge 1$. Proof: If not then write $y=\frac{x}{1-f(x)}$ giving $f(x)=2$, contradiction. Claim: $f(x)\ge 2$. Proof: If not then take some $f(y)<2$ and notice that $f(x)\cdot \frac{f(y)}{2}=f(x+yf(x))$. Fix $y$ and continue replacing $x\to x+yf(x)$ to derive some infinite geometric series with ratio $\frac{f(y)}{2}<1$, a contradiction to the first claim. Claim: $f$ is nondecreasing. Proof: Consider $a<b$. Choose $x,y$ with $x=a$ and $x+yf(x)=b$. Then $f(b)\ge f(a)$. Claim: If there exists $f(x_0)=2$, then $f\equiv 2$. Proof: Write $x_{i+1}=x_i+x_if(x_i)$. Verify that $f(x_i)=2$ always. Hence $x_{i+1}=3x_i$ which is unbounded. So there are arbitrarily large values which output $2$. Hence $f\equiv 2$ by nondecreasing. Claim: $f>2$ cannot hold everywhere. Proof: Verify that $f(x+yf(x))=f(y+xf(y))$ by swapping $x$ and $y$. Suppose we have found values $x,y$ where $x+yf(x)<y+xf(y)$. The interval $[x+yf(x),y+xf(y)]$ is constant. Now fix $x$ and increase $y\to y'$ where $x+y'f(x)=y+xf(y)$. The key is that the interval $[x+y'f(x),y'+xf(y')]$ is constant, hence $[x+yf(x),y'+xf(y')]$ is constant. If we continue applying the increment $y\to y'$ infinitely many times, then I claim that the upper bound of the interval is unbounded. The proof is quite simple: assume the current interval has size $s$. Then we increase $y$ by exactly $\frac{s}{f(x)}$, and the size of the interval increases by at least $\frac{s}{f(x)}$. Hence the new interval has size at least \[s\cdot \left(1+\frac{1}{f(x)}\right)\]which is unbounded. Then all large values are constant, hence take such $x,y$ and now $f(x)=f(y)=2$, a contradiction. It remains to show that $x+yf(x)=y+xf(y)$ does not hold everywhere. This is left as an exercise to the reader (happy new year, it's 12:10 AM and an hour past bedtime, my parents are yelling, will update if I can) Hi I'm back. If this holds always then \[xf(y)-x=yf(x)-y\implies x(f(y)-1)=y(f(x)-1)\]hence $\frac{x}{f(x)-1}=c$ so then $f(x)=\frac{x}{c}+1$. Taking $x\to 0$ contradicts $f(x)\ge 2$. Yay!

The solution is $f\equiv 2$. Now, if for some $x_0$, we have $f(x_0) < 1$, then $P\left(x_0,\dfrac{x_0}{1-f(x_0)}\right) \implies f(x_0) f\left(\dfrac{x_0}{1-f(x_0)}\right) = 2f\left(\dfrac{x_0 - x_0f(x_0) + x_0f(x_0)}{1-f(x_0)}\right) \implies f(x_0) = 2$, contradiction. Therefore $f \ge 1$. Now $P(x,x)$ gives that $f(x)^2 = 2f(x+xf(x)) \ge 2\cdot 1 \implies f(x) \ge \sqrt{2 \cdot 1}$. Now define $a_{n+1} = \sqrt{2a_n}$ with $a_1 = 1$. So, $a_n \le 2 \implies a_{n+1} = \sqrt{2a_n} \le \sqrt{2\cdot 2} = 2$. Also, $a_{n+1} = \sqrt{2a_n} \ge \sqrt{a_n \cdot a_n} = a_n$. Therefore by Monotone Convergence Theorem, we get that $a_n$ converges and that the limit is $=2$. Therefore we can repeat the step we did to go from $f \ge 1$ to $ f\ge \sqrt{2\cdot 1}$ repeatedly to get the RHS arbitrary close to $2$ using which we conclude $f \ge 2$. Now $P\left(x,\dfrac{\varepsilon}{f(x)}\right) \implies 2f(x+\varepsilon) = f(x)f\left(\dfrac{\varepsilon}{f(x)}\right) \ge 2f(x) \implies f(x+\varepsilon) \ge f(x)$. So $f$ is non-decreasing. Thus we can conclude that $\lim_{x\rightarrow 0^+} f(x)$ exists say to $a$. So $a \ge 2$. Firstly note that $\displaystyle\lim_{x \rightarrow 0^+} (x+xf(x)) = (0 + 0 \cdot a) = 0$. Now note that, \[ \displaystyle\lim_{x \rightarrow 0^+} f(x)^2 = \displaystyle\lim_{x \rightarrow 0^+} 2f(x+xf(x)) \implies a^2 = 2a \implies a = 2. \] Now fix $x_0$ arbitrarily. $f(x_0)f(y) = 2f(x_0 + yf(x_0)) \implies \displaystyle\lim_{y \rightarrow 0^+} f(x_0)f(y) = 2f(x_0)$. Therefore, $f$ is continuous at $x_0$ by choosing $\dfrac{y}{f(x_0)} \rightarrow 0$. Now fix $y$, \begin{align*} &\displaystyle\lim_{x \rightarrow 0^+} f(x)f(y) = 2 \displaystyle\lim_{x \rightarrow 0^+} f(x + yf(x))\\ \implies &2f(y) = 2f(2y) \implies f(y) = f\left(\frac{y}{2}\right) = f\left(\frac{y}{4}\right) = \cdots = f\left(\frac{y}{2^n}\right) .\end{align*} Now taking $n \rightarrow \infty$, we get that $f$ is constant using the continuity.

The answer is $f(x)=2$ for all $x\in \mathbb{R}_{>0}$ . It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. We first prove the following. Claim : $f(x)\geq 2$ for all $x\in \mathbb{R}_{>0}$. Proof : First, assume there exists $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha) <1$. Then, plugging in $x=\alpha$ and $y= \frac{\alpha}{1-f(\alpha)}$ gives \begin{align*} f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\alpha + \frac{\alpha}{1-f(\alpha)}\cdot f(\alpha)\right) \\ f(\alpha)f\left(\frac{\alpha}{1-f(\alpha)}\right) &= 2f\left(\frac{\alpha}{1-f(\alpha)}\right) \end{align*}from which it follows that $f(\alpha)=2$ which is a very clear contradiction. Thus, it follows that $f(x) \geq 1$ for all $x\in \mathbb{R}_{>0}$. Now, note that we have \[f(x)^2 = 2f(x+xf(x))\geq 2\]from which it follows that $f(x) \geq \sqrt{2}$ for all $x\in \mathbb{R}_{>0}$. Similarly, a straightforward induction gives, \[f(x) \geq 2^{\sum{\frac{1}{2^i}}}=2\]for all $x\in \mathbb{R}_{>0}$, which proves the claim. Now, we show that there exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$. First note that swapping $x$ and $y$ gives us that \[f(x+yf(x))=f(y+xf(y))\]for all positive real numbers $x$ and $y$. Then, say $x+yf(x)=y+xf(y)$ for all $x,y \in\mathbb{R}_{>0}$. Then, plugging in $y=1$ gives $f(x)=cx+1$ for some constant $c$ for all positive reals $x$ which is clearly false for any $c\in \mathbb{R}_{>0}$ upon substitution. This means, there exists $x_0,y_0 \in \mathbb{R}_{>0}$ such that $x_0+y_0f(x_0) \neq y_0+x_0f(y_0)$. WLOG, assume that $x_0+y_0f(x_0)>y_0+x_0f(y_0)$. Then, let $b= y_0+x_0f(y_0)$ and $a=\frac{x_0+y_0f(x_0) - (y_0+x_0f(y_0))}{f(y_0+x_0f(y_0))}>0$. Then, plugging in $x=b$ and $y=a$ yields, \[f(b)f(a)=2(f(b+af(b)))=2f(x_0+y_0f(x_0)) = 2f(b)\]from which it follows that $f(a)=2$. Thus, there indeed exists $r\in \mathbb{R}_{>0}$ such that $f(r)=2$. We now have our final key claim. Claim : $f$ is non-decreasing. Proof : Simply note that \[2f(x+yf(x))=f(x)f(y)\geq 2f(x)\]since $f\geq 2$. Thus, $f(x) \leq f(x+yf(x))$ for all $x,y \in\mathbb{R}_{>0}$. Thus, it follows that $f(x)\leq f(x+\epsilon)$ for any $x, \epsilon \in \mathbb{R}_{>0}$ (by varying $y$) which implies that $f$ is non-decreasing as claimed. Now, consider $\alpha \in \mathbb{R}_{>0}$ such that $f(\alpha)=2$. Plugging in $x=y=\alpha$ gives \[4=f(\alpha)^2=2f(\alpha+\alpha f(\alpha))=2f(3\alpha)\]which implies that $f(3\alpha)=2$. Thus, it follows by a straightforward induction that $f(3^ra)=2$ for all $r\in \mathbb{N}$, from which it follows that there exists arbitrarily large values of $x\in\mathbb{R}_{>0}$ such that $f(x)=2$. Putting this together with the non-decreasing nature of $f$ it follows that $f$ is constant 2, as desired.

Let $P(x,y)$ be the assertion. Claim: $f(x)\ge 1$. Proof: Suppose otherwise. If there exists $a$ where $f(a)<1$ then we can find \[y=a+yf(a)\]hence $P(a,y)$ yields $f(a)=2$, a contradiction. Claim: $f(x)\ge 2$. Proof: Suppose otherwise. Let $S$ be the range of $f$. From $P(x,x)$ with $v=f(x)<2$ we find \[v\in S\implies \frac{v^2}{2}\in S\]hence we can generate a sequence $v_0,\dots,v_n\dots$ and note that \[v_{i+1}\le \frac{v_0}{2}\cdot v_i\]thus values become arbitrarily small. Hence at some point values are less than $1$, violating the first claim. Claim: $f$ is nondecreasing. Proof: Take any $a$ and $b$ with $a<b$. Notice we can always choose $x$ and $y$ such that \[a=x\]\[b=x+yf(x).\]Then since $f(y)\ge 2$ we find $f(a)\le f(b)$. Claim: Either $f$ is increasing or $f\equiv 2$. Proof: Suppose there exist $a$ and $b$ such that $f(a)=f(b)$. Then $f$ is constant in the interval $[a,b]$. At this point take $P(a,y)$ such that \[a+yf(a)\in (a,b].\]For any such $y$ we have $f(y)=2$. Hence there exists an interval where $f\equiv 2$. Take $x$ in this interval and $P(x,x)$ implies $f(3x)=2$. Hence we can obtain arbitrarily large values outputting $2$, and so $f\equiv 2$ everywhere. Claim: $f$ increasing is not possible. Proof: Swap $x$ and $y$. Then \[f(x+yf(x))=f(y+xf(y))\implies x+yf(x)=y+xf(y)\implies \frac{f(x)-1}{x}=\frac{f(y)-1}{y}\]hence $f(x)=cx+1$ for some $c$. However taking $x\to 0$ is a contradiction as we would have $f(x)\to 1$ which is less than $2$. To conclude, we have one solution (which clearly works): $\boxed{f\equiv 2}$.

Very instructive problem! This reviews many useful techniques in FE. The answer is $f(x)=2~~\forall x \in \mathbb{R}^+$ which is clearly satisfy the problem condition. We’ll show that this is the only solution. Denote $P(x,y)$ by the condition $f(x)f(y)=2f(x+yf(x))~~ \forall x,y \in \mathbb{R}^+$ Claim 1: $f(x)\geq 1~~ \forall x \in \mathbb{R}^+$ Proof: FTSOC, suppose there exist $t \in \mathbb{R}^+$ s.t. $f(t)<1$ Consider $P(t,\frac{t}{1-f(t)})$, it follows that $f(t)=2$ which leads to contradiction. $\square$ Claim 2: $f(x)\geq 2~~ \forall x \in \mathbb{R}^{+}$ Proof: Define a sequence ${a_n}$ s.t. $a_1=1$ and $a_{n+1}=\sqrt{2a_{n}}~~ \forall n \in \mathbb{N}$ It is easy to see that $lim_{n\to \infty}a_n=2$ and $a_n<2~~ \forall n \in \mathbb{N}~~(\clubsuit)$ We prove that $f(x)\geq a_n ~~\forall x \in \mathbb{R}^+ , n \in \mathbb{N}$ by induction on $n$. The base case is proven in Claim 1 Suppose that $f(x)\geq a_m~~ \forall x \in \mathbb{R}^+$ Consider $P(x,x)$, $f(x)^2=2f(x+xf(x))\geq 2a_m~~ \forall x \in \mathbb{R}^+$ $$\implies f(x)\geq \sqrt{2a_m}=a_{m+1}~~\forall x \in \mathbb{R}^+ $$So, $f(x)\geq a_n ~~\forall x \in \mathbb{R}^+ , n \in \mathbb{N}$ Combine this with $(\clubsuit)$, $f(x)\geq 2~~ \forall x \in \mathbb{R}^{+}$, as desired. $\square$ Claim 3:$f$ is non decreasing Proof: Let $a<b$ be any two distinct positive reals Consider $P(a,\frac{b-a}{f(a)})$, $f(a)f(\frac{b-a}{f(a)})=2f(b)$ Since $f(x)\geq 2$ for all $x$, it follows that $f(a)\leq f(b)~~\square$ Now we divide into 2 cases Case 1) There exist $c \in \mathbb{R}^+$ such that $f(c)=2$ Note that if we have $f(x)=2$, it implies that $f(x+xf(x))=2$ (This followed from $P(x,x)$) We can easily prove that $f(3^nc)=2 \forall n \in \mathbb{N} $ by induction on $n$ Combine this result with Claim 2 and Claim 3, we conclude that $f(x)=2~~\forall x \in \mathbb{R}^+$ Case 2) $f(x)>2~~ \forall x \in \mathbb{R}^+$ Claim: $f$ is strictly increasing Proof: We can easily copied the proof in Claim 3. $\square$ This implies that $f$ is injective. The trick is to swap the variable in the problem assertion. Consider $P(x,y)$ and $P(y,x)$, $f(x+yf(x))=f(y+xf(y)) \forall x,y \in \mathbb{R}^+$ By injectivity, $x+yf(x)=y+xf(y) \forall x,y \in \mathbb{R}^+$ $\implies \frac{f(x)-1}{x}$ is a constant $\forall x \in \mathbb{R}^+$ So, $f(x)=cx+1 \forall x \in \mathbb{R}^+$ for some constant $c \in \mathbb{R}^+$ which can be easily checked that it fails. $\blacksquare$

The answer is $f(x) = \boxed{2}$, which works. Denote the given assertion as $P(x,y)$. We will solve this problem through a series of claims: Claim 1: $f(x) \ge 1$ for all $x \in \mathbb{R}^+$ Proof: FTSOC suppose that $x$ satisfies $f(x)<1$. Plugging in $P(x, \tfrac{x}{1-f(x)})$ yields \[f(x)f\left(\frac{x}{1-f(x)}\right) = 2f \left(\frac{x}{1-f(x)}\right) \implies f(x) = 2,\] a contradiction. $\square$ Claim 2: $f(x) \ge 2$ for all $x \in \mathbb{R}^+$ Proof: $P(x,x)$ yields \[f(x)^2 = 2f(x+f(x)) \ge 2.\] Hence, $f(x) \ge \sqrt{2}$. In fact, we can show that \[f(x) \ge \lim_{k \to \infty} 2^{\tfrac{2^k-1}{2^k}} = 2\] using an inductive process on $P(x,x)$, as desired. $\square$ Claim 3: $f$ is nondecreasing Proof: Note that \[2f(x+yf(x)) = f(x)f(y) \ge 2f(x),\] which means $f(x) \le f(x+yf(x))$. Take $y = \varepsilon$ to be an arbitrarily small number, so that $f(x) \le f(x+\varepsilon)$ and $x < x+ \varepsilon$. $\square$ Claim 4: If there exists any value $x \in \mathbb{R}^+$ such that $f(x) = 2$, then $f \equiv 2$. Proof: If $f(x) = 2$, then \[f(3x) = f(x+xf(x)) = \frac{f(x)^2}{2} = 2.\] Thus, the set of $x$ such that $f(x)=2$ is unbounded, and Claim 3 finishes. $\square$ Claim 5: $f$ is not injective. Proof: Assume FTSOC that $f$ is injective. Then, comparing $P(x,y)$ and $P(y,x)$ gives \[f(x+yf(x)) = f(y+xf(y)) \implies x+yf(x) = y+xf(y)\]\[\implies x(1-f(y)) = y(1-f(x)) = k.\] Rearraging yields $f(y) = 1-\tfrac{k}{x}$, which implies $f(k) = 0$, contradiction. $\square$ Claim 6: $f>2$ cannot always hold. Proof: Suppose that $x$ and $y$ satisfy $x+yf(x) < y+xf(y)$. Such values must exist due to Claim 5 preventing $x+yf(x) = y+xf(y)$. Note that all values in the interval $[x+yf(x), y+xf(y)]$ are constant. The trick is to raise $y \to y'$ such that $x+y'f(x) = y+xf(y) \le y'+xf(y')$. Hence, we can expand the constant interval to $[x+yf(x), y'+xf(y')]$. It suffices to show that increasing $y$ infinitely many times causes the interval to be unbounded. Let $\ell$ be the size of the current interval. Each time, we increase $y$ by $\tfrac{\ell}{f(x)}$, which increases the size of the next interval by at least $\tfrac{\ell}{f(x)}$. Hence, the new interval at least has size \[\ell\left(1+\frac{1}{f(x)} \right),\] which is evidently unbounded. $\square$ Claims $4$ and $6$ imply that $f$ is constant, which gives our solution set.

The answer is the constant $2$ only, it is easy to check that this works. Let $P(x,y)$ denote the assertion. First, observe is $f(x)$ is below $1$ at any point, we can find $y$ such that $x + yf(x) = y$, plugging in such $y$ forces $f(x) = 2$, thus $f$ can never be below $1$ (since we already said $f(x)$ was below $1$). If $f(x)$ is below $2$ at any point, we can always see there exists a value of $k$ such that $f(k) = \frac 12 f(x)^2$, which eventually goes below $1$ (it is bounded by a geometric series with common ratio less than $1$), so this returns to the case where $f(x)$ is below $1$ at some point. So we know $f \ge 2$. Claim: $f$ is $2$ at some point. If $f$ is always above $2$, we see $f(x) \frac{f(y)}{2} =f(x +yf(x))$, so for all $a > x$, setting $y = \frac{a - x}{f(x)}$ forces $f(a) > f(x)$. Now take $2f(x + yf(x)) = f(x)f(y) = f(y)f(x) = 2f(y +xf(y))$, since $f$ is increasing we know $f$ is injective, this gives $x + yf(x) = xf(y) + y$, for sufficiently small $y$ the left side is below $2x$ and the right side is greater, which is a contradiction, so this is impossible. Claim: $f$ is periodic. Proof: Take $c$ with $f(c) = 2$, then we get $P(x,c)$ forces $f(x) = f(x + 2c)$. Claim: $f(x) = 2$ for all $x$. Proof: We first show $f(x) = 1$ or $f(x) = 2$ for all $x$. Assume $f(x) \neq 1$, we show $f(x) = 2$. By varying $y$, we can find some value of $y$ such that there exists an integer $k$ with $x + yf(x) - y = 2kc$, this forces $f(y) = f(x + f(y)) $, so $P(x,y)$ gives $f(x) = 2$. Now take some value with $f(x) = 1$, then $P(x,x)$ gives $f(x + xf(x)) = \frac 12$, which is impossible, so such a value cannot exist, giving the desired conclusion.

I hope this is right Denote by $P(x, y)$ the given assertion. Claim 1: $f$ is monotonically increasing. Proof: take two numbers $a < b$ and assume FTSOC $f(a) > f(b)$. Let $r$ be the solution of the equation $a + rf(a) = b+ rf(b)$(it exists because of the way we ordered these two numbers). $P(a, r): f(a)f(r)=2f(a + rf(a))$ $P(b, r): f(b)f(r)=2f(b + rf(b))$ So $f(a)=f(b)$, contradiction. $\square$ Now we will prove that $f$ is not injective( i.e strictly increasing) Suppose FTSOC that $f$ is injective. Swapping $x$ and $y$ in the original equation gives: $x + yf(x) = y + xf(y), \forall x > 0$ $ \Rightarrow \frac{f(x)-1}{x} = k, \forall x > 0$ Plugging this in the original equation, we get a contradiction. $\square$ Since $f$ is not injective, we can find $a < b$ such that $f(a) = f(b) = c$. By monotonicity, we get that $\forall x \in [a, b], f(x) = c$. $P(a, \frac{b-a}{c}): f(a)f(\frac{b-a}{c})=2f(b)$ so we can find $z \in \mathbb{R}$ such that $f(z)=2$ $P(z, x): f(x) = f(2x + 2), \forall x \in \mathbb{R}$ Set $x=b$ $\Rightarrow f(b) = f(2b+2) =c$. Using monotonicity again, we get that $\forall x \in [b,2b+2], f(x) = c$ Similarily, $\forall x > b, f(x) =c$ Finally, $P(a, x): cf(x) = 2(a + xc)$. Since the second argument $>a$ we get that $f(x) = 2 \forall x \in \mathbb{R} $ which is clearly the only solution $\blacksquare$

We claim $f\equiv 2$ is the only solution, which clearly works. First, if $f(x)<1$ then $P(x,\tfrac x{1-f(x)})$ gives $f(x)=2$. Thus $f(x)\ge 1$. Next, we prove by induction that $f(x)\ge 2^{1-2^{-n}}$. From the above, the base case $n=0$ works, and for the inductive step $P(x,x)$ gives $f(x)^2=2f(x+xf(x))\ge 2^{2-2^{-n}}$ implying $f(x)\ge 2^{1-2^{-(n+1)}}$, completing the induction. Taking $n$ arbitrarily large implies $f(x)\ge 2$ for all $x$. Next we show $f$ is constant or strictly increasing. From $f(y)\ge 2$ we get $f(x)\le f(x+yf(x))$. If $f$ is not strictly increasing we can choose some $x$ and some $\varepsilon$ for which $f(x)=f(x+kf(x))$ for $0<k\le\varepsilon$, and by equality case $f(k)=2$ for $0<k\le\varepsilon$. Now $P(x+\varepsilon f(x),k)$ implies $f(x)=f(x+(k+\varepsilon)f(x))$, so by equality case again $f(k)=2$ for $0<k\le 2\varepsilon$. Repeating this, $f(k)=2$ for all $k>0$. Now if $f$ is strictly increasing, it is injective. Taking $P(x,1)$ and $P(1,x)$ implies $x+f(x)=1+xf(1)$, which fails for $x$ small enough since $f(x)\ge 2$. Thus $f\equiv 2$ is the only possible solution.