We have $f(x)>f(f^2(x)+y)$. We'll derive a LOT of information from this alone. Equivalently, this says: \[ C>f^2(x) \implies f(C)<f(x) \quad \forall C,x. \qquad (\spadesuit)\]If $x>f^2(x)$, then $C=x$ in $(\spadesuit)$ gives $f(x)<f(x)$, contradiction. Hence $\boxed{f^2(x) \ge x}$. So $f^2(x) \le f^4(x) \le f^6(x) \le \cdots$ by iterating. Also $f(x) \le f^3(x) \le f^5(x) \le \cdots$. The contrapositive of $(\spadesuit)$ is very useful: \[ f(C) \ge f(x) \implies C \le f^2(x) \quad \forall C,x. \qquad (\heartsuit)\]Since $f^5(x)\ge f(x)$, plugging in $C=f^4(x)$ into $(\heartsuit)$ gives $f^4(x) \le f^2(x)$. But recall $f^4(x) \ge f^2(x)$ too, so $\boxed{f^2(x)=f^4(x)}$. Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$ and $b>a$. Now we utilize the entire FE. Plug in $x=f^2(n)$ into the original FE for some arbitrary real $n$: \begin{align*} f^3(n) &= f(f^4(n)+y) + f(f^2(n)\cdot f(y))\cdot f(f^2(n)+y) \\ &= f(f^2(n)+y) \cdot [1+f(f^2(n)\cdot f(y))]. \end{align*}Plug in $y=a$ and $y=b$ into the above and compare. Then we get $f(f^2(n) + a) = f(f^2(n) + b)$. Now the original FE with $x=n$ reads: \[ f(n) = f(f^2(n) + y ) + f(nf(y))f(n+y). \]Plugging in $y=a$ and $y=b$ and comparing gives $f(n+a)=f(n+b)$. This holds for arbitrary $n$, so $f(x+a)=f(x+b)$ for all $x$. Now iterating and chaining this, we get $f(x+a) = f(x+B)$ for arbitrarily large $B$ since $b>a$. Take large enough $B$ for which $x+B>f^2(x+a)$. Then by $(\spadesuit)$, $f(x+B) < f(x+a)$, contradiction. Hence $b=a$. $\blacksquare$ Remember $f^2(x)=f^4(x)$. So now by injectivity, $\boxed{f^2(x)=x}$. Now $(\heartsuit)$ reads to give $f$ decreasing. The FE is now: $f(x)=f(x+y)[1+f(xf(y))]$. Plugging in $x=1$ gives $f(y+1)=f(1)/(y+1)$. So $f(x)=C/x$ for $x>1$ for some constant $C$. So since $f$ is an involution, $f(x)=C/x$ for $x<C$ too. Now we want $f$ for sub-1 inputs. Fix $x\le 1$, and plug in $y=100$ into the new FE, and we get $f(x)=C/x$. Now we can confirm $f(x)=C/x$ works for all $x>0$, and it must be the solution form.

The answer is $f(x)\equiv \frac ax$ which clearly works. Let $P(x,y)$ denote the assertion in the question. Claim 1 $x\le f(f(x))$. Proof: Assume not, then $P(x,x-f(f(x)))$ gives $f(x)=f(x)$+smth nonzero, contradiction. Now, notice $f(x)\le f(f(f(x)))$ but $f(x)>f(f(f(x))+y)$ for any $y\in \mathbb{R}^+$ Claim 2: If $f(f(x))>f(f(y))$ then $f(x)<f(y)$. Proof: $P(y,f(f(x))-f(f(y)))$ gives $f(y)>f(f(f(x)))\ge f(x)$. Claim 3: $f(f(x))=f(f(f(f(x))))$ for all $x\in \mathbb{R}^+$ Proof: Assume not. Then $f(f(x))<f(f(f(f(x))))$ and $f(x)<f(f(f(x)))$. Let $A=f(f(x))$, then we have $f(x)<f(A)$ and $f(f(x))<f(f(A))$, but $f(f(x))<f(f(A))$ implies $f(x)>f(A)$, absurd. Claim 4: $f(x)<\frac{f(1)}{x-1}$ Proof: $P(1,x-1)$. Claim 5:$f(x)\equiv f(f(f(x)))$ for all $x\ge 1$ Let $B=f(f(x))$. We are given $f(f(x))=f(f(B))$ by claim 3. Assume $f(x)<f(B)$ since $f(x)\le f(f(f(x)))$. This also means $1\le x<B.$ Then $P(x,y), P(B,y)$ yields $$f(x)=f(f(f(x))+y)+f(x+y)f(xf(y))$$ $$f(B)=f(f(f(B))+y)+f(B+y)f(Bf(y))$$ Let $C=f(B)-f(x)$. Plugging in $y=f(t)$ gives $f(B)-f(x)< f(B+f(t))f(Bf(f(t))) <\frac{f(1)}{B-1} \cdot \frac{f(1)}{Bf(f(t))-1} \le \frac{f(1)}{B-1} \cdot \frac{f(1)}{Bt-1}$ which tends to 0 as $t$ tends to infinity. Claim 6 if $x\ge 1$ then $f(f(x))> f(f(x)+y)$ for any $y\in \mathbb{R}^+$. Proof: $P(f(x),y)$ gives $f(f(x))=f(f(x)+y)+f(f(x)+y)f(f(x)f(y))$. Also, from $P(f(f(x)),y)-P(x,y)$ we get for all $x\in \mathbb{R}_{\ge 1}, y\in \mathbb{R}^+$, $$f(f(f(x))+y)f(f(f(x))f(y)) = f(x+y)f(xf(y))$$is an identity, call $Q(x,y)$. Claim 7 $f(f(1))=1$. If $x=1,$ and $f(f(1))>1$ we have $f(f(y))>f(f(f(1)) \cdot f(y))$ for all $y\in \mathbb{R}_{\ge 1}$ because $f(f(1))\cdot f(y)$ and claim 6. Therefore, we select $y$ such that $1+y=f(f(z))$ where $z\ge 2$. We can see that $f(1+y)=f(f(f(z)))>f(k)$ for any $k>f(f(z))$. Therefore, $f(1+y)>f(f(f(1))+y)$. Multiplying, $f(f(y))f(1+y) > f(f(f(1))+y) f(f(f(1))f(y))$ for this $y\in \mathbb{R}_{\ge 1}$ Claim 8 $f(1+y)=f(1+f(f(y)))$ for all $y\ge 1$ $P(1,y)$ yields $f(1)=f(1+y)(1+f(f(y)))$ $P(1,f(y))$ yields $f(1)=f(1+f(y))(1+f(y))$ where $y\ge 1$. Thus, for $y\ge 1, f(1+f(y))=\frac{f(1)}{1+f(y)}$ In particular, $f(1+f(f(y)))=\frac{f(1)}{1+f(f(y))}=f(1+y)$ Claim 9 Let $a=f(1)$, then for all $x\ge 1$, $f(x+a)=f(f(f(x))+a)$. Proof: $Q(x,a)$ gives $f(f(f(x))+a)f(f(f(x)))=f(x+a)f(x)$ since $f(a)=1$. Since $f(f(f(x)))=f(x)$, the conclusion follows. Now, $x$ denotes a real that is at least 1. Claim 10 $f(x)f(af(x))=1$ for all $x\ge 1$ $P(x,a)$ gives $f(x)=f( f(f(x))+a)+f(x+a)f(x)=f(x+a)(1+f(x))$. Therefore, $f(x+a)=\frac{f(x)}{1+f(x)}$ $P(a,x)$ gives $f(a)=f(f(f(a))+x)+f(a+x)f(af(x))=f(a+x)(1+f(af(x))$ Therefore, $1=\frac{f(x)}{1+f(x)}(1+f(af(x)))$ $1+f(x)=f(x)(1+f(af(x))$ $1=f(x)f(af(x))$ for all $x\ge 1$ Claim 11 For all $x\ge 1$, $f(x)f(f(x))=a$. $P(1,x)$ gives $a=f(1+x)+f(1+x)f(f(x))$. Therefore, for all $x\ge 0$, $f(1+x)=\frac{a}{1+f(f(x))}$ In particular, $f(1+f(x))=\frac{a}{1+f(x)}$ if $x\ge 1$. Now, $P(f(x),1)$ where $x\ge 1$ gives $f(f(x))=f(f(x)+1)+f(f(x)+1)f(af(x))$ $f(f(x))=\frac{a}{1+f(x)} (1+f(af(x)))$. By Claim 10, if $x\ge 1$ then $f(af(x))=\frac{1}{f(x)}$ Therefore, $f(f(x))=\frac{a}{1+f(x)} (1+\frac{1}{f(x)}) = \frac{a}{1+f(x)} \frac{1+f(x)}{f(x)}$, so $f(f(x))f(x)=a$. In particular, $f(x)\le \frac ax$ By Claim 11, for all $x\ge 1$, we have $f(x)=\frac{a}{f(f(x))}\le \frac ax$ and if $f(x)=\frac ax$ then it follows that $f(f(x))=x$. This means $f(x)=\frac ax$ for a dense set in $\mathbb{R}_{\ge 1}$, call $S$. This set $S$ satisfies for all $t\in S, f(f(t))=t$. We note this set $S$ is also $Im(f)=Im(f(f(x))$. Indeed, if $f(t)=u$ then $uf(u)=a$ and if $uf(u)=a$ then $u=f(\frac au)$ Claim 12 $f(xf(y))=f(f(f(x))f(y))$ By $Q(x,f(y))$, we have $f(f(f(x))+f(y))f(f(f(x))f(f(y)))=f(x+f(y))f(xf(f(y)))$. By $P(f(x),y)-P(f(x),f(f(y))$, we can get $f(f(x)+y)=f(f(x)+f(f(y))) = \frac{f(f(x))}{1+f(f(x)f(y))}$. Therefore, $f(f(f(x))+f(y))=f(x+f(y))$, it follows that $f(xf(f(y)))=f(f(f(x))f(f(y)))$. Since the range of $f(f(y))$ is precisely the range of $f(y)$, it follows that $f(xf(y))=f(f(f(x))f(y))$ Therefore, $f(xf(y))=\frac{f(x)}{f(y)}=\frac{f(f(f(x)))}{f(y)}=f(f(f(x))f(y))$. Now, $Q(x,y)$ gives $f(x+y)=f(f(f(x))+y)$ for all $x\ge 1, y\in \mathbb{R}^+$. Picking $y=f(f(z))-x$ gives a contradicting $f(t)>f(u)$ if $u>t$ and $t\in S$. Therefore, it follows that $f(f(x))=x$ for all $x\ge 1$. Since $S$ is closed under inverses, $S=\mathbb{R}^+$, as desired.

Let $P(x,y)$ denotes the equation.

@below Yep. On another note, I didn't see that the method in claim 2 works. That's quite a good job, I suppose

wow I made more progress than you during the test. I proved $f(x)=f^3(x)$ :O @above I didn't realize injectivity works that would've cut down my solution length to half

My problem! The main trickery is basically if we have the condition $f(x)>f(f(f(x))+y)$, then surprisingly we can prove $f^{(4)}(x)=f^{(2)}(x)$!!! The two solutions above more or less do the same thing, but I didn't notice that there's some way around to get $f^{(3)}(x)=f(x)$ (as you can see, I have the trick in mind when composing this problem so it became extremely hard for me to see other solutions lol). I think the solution in #4 is really nice, but for those who only got $f^{(4)}(x)=f^{(2)}(x)$, here is one way to get injectivity: If $a>b$ but $f(a)=f(b)$, consider $P(f^{(2)}(x), a)$ and $P(f^{(2)}(x), b)$. By comparing them, we get $f(f^{(2)}(x)+a)=f(f^{(2)}(x)+b)$. Now plugging this in back to $P(x,a), P(x,b)$, and we get $f(x+a)=f(x+b)$. But periodicity is clearly absurd---for example, take $x=f^{(2)}(a)-b$ (which is positive as $f^{(2)}(a)\geq a$. Therefore $f$ is injective. (Credit to FEcreator for this clean injectivity argument.) From here, one can really do anything to finish the solution.

We claim $f(x) = \frac cx$ for some constant $c>0$. It is clear that $f$ works. Now consider some working function $f$. Let $P(x,y)$ denote the condition. Abbreviate $f(f(f(x)))$ to $f^3(x)$ etc. for convenience. Observe that for $t>f(f(x))$, taking $P(x,t-f(f(x)))$ yields $f(x)>f(t)$. In particular, we must have $f(f(x)) \ge x$ for all $x$ because we otherwise get $f(x)> f(x)$, absurd. Claim: For all $x$, we have $f^4(x) = f^2(x)$. Solution: Suppose otherwise. Then for some $x$, we have $f^4(x) > f^2(x)$. Then we get $f^5(x) < f(x)$, which is absurd because $f^3(x)\ge f(x)$ and $f^5(x)\ge f^3(x)$, so we are done. $\fbox{}$ Taking $P(f^2(x),y)$ yields \[f^3(x) = f(f^2(x)+y)+f(f^2(x)f(y))f(f^2(x)+y) = f(f^2(x)+y)\cdot [1+f(f^2(x)f(y))]\qquad (\diamondsuit).\]Observe that $P(1,y)$ yields \[f(1) = f(f(f(1))+y)+f(f(y))f(1+y)\ge f(f(f(1))+y)+yf(1+y).\]Hence we have the bound \[\frac{f(1)}{y} > f(1+y) \qquad (\clubsuit).\] Claim: $f$ is injective. Solution: Suppose otherwise. Then there exist some $a<b$ with $f(a)=f(b)$. By $\diamondsuit$, this implies that for all $x$ we have $f(f^2(x)+a)=f(f^2(x)+b)$. In turn, taking $P(x,a)$ and $P(x,b)$ yields $f(xf(a))f(x+a) = f(xf(b))f(x+b)$, so $f(x+a)=f(x+b)$. Let $k$ be the minimum positive integer such that $f(1)10^{-k} < f(a+1)$. Define $N = \lceil \frac{10^k}{b-a}\rceil$ to yield \[f(1)10^{-k}<f(a+1) = f(a+1+(b-a)) = \cdots = f(a+1+N(b-a)) < \frac{f(1)}{N(b-a)} \le f(1)10^{-k}\]by $\clubsuit$, absurd. $\fbox{}$ Then $f^4(x) = f^2(x)$ implies $f^3(x) = f(x)$ and $f^2(x)=x$, so $f$ is surjective and the given equation rewrites as \[f(x)=f(x+y)+f(xf(y))f(x+y) = f(x+y)\cdot [1+f(xf(y))].\]Take $P(1,y)$ to yield $f(1) = f(1+y)\cdot [1+f(f(y))] = (1+y)f(1+y)$. Thus for $x > 1$, we have $f(x) = \frac 1x f(1)$. Then for $x>1$, $P(x,y)$ rewrites as \[\frac 1x f(1) = \frac{1}{x+y}f(1)\cdot [1+f(xf(y))],\]which rearranges to $\frac yx = f(xf(y))$. For $y>1$, this is $\frac yx = f(f(1)\cdot \frac xy)$. As all reals may be expressed as $\frac xy$ for some $x,y>1$, this means that $f(x)=\frac 1x f(1)$ for all $x$. This is equivalent to the stated solution set.

Let $\mathbb{R}_{>0}$ be the set of all positive real numbers. Find all functions $f:\mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that for all $x, y \in \mathbb{R}_{>0}$ we have $$f(x) = f (f(f(x)) + y) + f(xf(y))f(x + y).$$. Solved with hint from rama1728 (the solution set) Let $P(x,y)$ denote the given assertion. We claim the answer is $\boxed{f(x)=\frac{c}{x}}$ for some real constant $c$. It's easy to check that this works. We now prove it's the only solution. Claim 1: If $f$ is involution, we are done. Fixed proof thanks to jasperE3: Suppose $f(f(x))=x$ for all $x$. $P(1,x): f(1)=f(1+x)+xf(x+1)\implies f(x+1)=\frac{f(1)}{x+1}$. So $f(x)=\frac{f(1)}{x}$ for $x>1$. Let $f(1)=c$. Set $x,y>1$. Note that $x>1$ and $x+y>1$. Then \[f(x)=f(x+y)+f\left(\frac{cx}{y}\right)f(x+y)=f(x+y)\left(f\left(\frac{cx}{y}\right)+1\right).\] So $\frac{c}{x}=\frac{c}{x+y}\left(f\left(\frac{cx}{y}\right)+1\right)$. This implies $\frac{x+y}{x}=f\left(\frac{cx}{y}\right)+1$, so $f\left(\frac{cx}{y}\right)=\frac{y}{x}$, so $f(x)=\frac{c}{x}$ for all $x>0$. $\blacksquare$ Claim 2: $f(f(x))\ge x$ for all $x$. Proof: If, not for some $x$, then if we set $y=x-f(f(x))$, then $f(xf(y))f(x+y)=0$. $\blacksquare$ Claim 3: If $f(f(y))>f(f(x))$, then $f(x)>f(y)$. Proof:\[P(x,f(f(y))-f(f(x)): f(x)=f^3(y)+f(xf(f(f(y))-f(f(x))))f(x+f(f(y))-f(f(x)))>f(y)\]$\blacksquare$ Claim 4: If $f(x)>f(y)$, then $f(f(y))>f(f(x))$, Proof: $P(f(y),f(x)-f(y)): f(f(y))=f(f^3(y)+f(x)-f(y))+f(f(y)f(f(x)-f(y)))f(f(x))>f(f(x))$. $\blacksquare$ Claim 5: $f^4(x)= f^2(x)$. Proof: Suppose for some $k$, $f^4(k)>f^2(k)$. $P(k,f^4(k)-f^2(k)): f(k)=f^5(k)+smth>f^5(k)$, but also $f^5(k)\ge f^3(k)\ge k$, a contradiction. $\blacksquare$ Claim 6: $f^3(x)=f(x)$. Proof: If $f^3(x)>f(x)$, then $f^4(x)<f^2(x)$, not true. $\blacksquare$ Claim 7: If $f$ is injective, we are done. Proof: If $f$ is injective, since $f^3(x)=f(x)$, $f^2(x)=x$. $\blacksquare$ Claim 8: $f$ is injective. Proof: $P(f(f(x)),y): f(x)=(f(f(f(x))f(y))+1)f(f(f(x))+y)$ So comparing with the original FE gives $f(xf(y))f(x+y)=f(f(f(x))f(y))f(f(f(x))+y)$ (1) Looking at $P(f(f(x)),a)$ and $P(f(f(x)),b)$, we find that if $f(a)=f(b)$, then $f(f(f(x))+a)=f(f(f(x))+b)$. Now looking setting $y=a$ and $y=b$ in $(1)$, we find $f(x+a)=f(x+b)$. So for $x>a$, $f(x)=f(x+b-a)$. If $b\ne a$, $P(x,b-a): f(x)=f(x)+smth$, where $smth>0$, a contradiction.

Very tricky one, i love it . We claim that $f(x)=\frac{c}{x}$ where $c \in \mathbb R_{>0}$ works. Let $P(x,y)$ the assertion of the given F.E. Claim 1: $f(f(x)) \ge x$ Proof: Assume that there exists $x$ such that $f(f(x))<x$ then by $P(x,x-f(f(x)))$ $$0=f(xf(x-f(f(x))))f(2x-f(f(x))) \; \text{contradiction!!}$$Claim 2: $f^{2k}(x)=f(f(x)) \; \forall k \in \mathbb Z_{>0}$ Proof: Assume that there exists $k \in \mathbb Z_{>0}$ and $x \in \mathbb R_{>0}$ such that $f^{2k}(x)>f(f(x))$ then by $P(x,f^{2k}(x)-f(f(x)))$ and Claim 1 $$f(x)=f^{2k+1}(x)+f(xf(f^{2k}(x)-f(f(x))))f(x+f^{2k}(x)-f(f(x))) \ge f(x)+f(xf(f^{2k}(x)-f(f(x))))f(x+f^{2k}(x)-f(f(x))) \implies 0 \ge f(xf(f^{2k}(x)-f(f(x))))f(x+f^{2k}(x)-f(f(x))) \; \text{contradiction!!}$$So $f^{2k}(x) \le f(f(x))$ but using Claim 1 we get $f^{2k}(x) \ge f(f(x))$ so $f^{2k}(x)=f(f(x))$ for any $k \in \mathbb Z_{>0}$ as desired. Claim 3: $f$ is injective. Proof: Assume that there exists $f(a)=f(b)$ with $a>b$ then comparing $P(f(f(x)),a)$ with $P(f(f(x)),b)$ along with Claim 2 $$f(f(f(x))+a)=f(f(f(x))+b)$$Now conparing $P(x,a)$ with $P(x,b)$ we get $$f(x+a)=f(x+b) \implies f(x)=f(x+a-b) \; \forall x \in \mathbb R_{>b}$$Now by $P(f(x),a-b)$ and Claim 2 $$1=1+f(f(x)f(a-b)) \implies f(f(x)f(a-b))=0 \; \text{contradiction!!}$$Hence $f$ is injective as desired. Finishing: By injectivity on Claim 2 we get $f(f(x))=x$ so $f$ is an involution and now we re-write the F.E. as $$f(x)=(1+f(xf(y)))f(x+y)$$Call $Q(x,y)$ the assertion of this F.E., then using $Q(1,x-1)$ where $x>1$ $$f(1)=xf(x) \implies f(x)=\frac{f(1)}{x} \; \forall x \in \mathbb R_{>1}$$Now set $x>1$ and use $Q \left(x, f \left(\frac{1}{x} \right) \right)$ $$1+\frac{f\left(\frac{1}{x} \right)}{x}=1+f(1) \implies f \left(\frac{1}{x} \right)=xf(1) \; \forall x \in \mathbb R_{>1}$$Now make the replace $t=\frac{1}{x}$ so here clearly $t$ can take all the values on $[0,1]$ so we got $f(t)=\frac{f(1)}{t} \; \forall t \in [0,1]$ So now let $f(1)=c$ and along with the other result we got $$\boxed{f(x)=\frac{c}{x} \; \forall x \in \mathbb R_{>0}}$$so we are done

I still can't solve it rip.

Serena_Xu wrote: I still can't solve it rip. I'm pretty sure u will soon

Solved with p_square We claim the only solutions are $f(x) = \frac{c}{x}$ for some $c \in \mathbb{R}^+$, which can be checked to work. Let $P(x,y)$ denote the given assertion. Claim 1: $f^4(x) = f^2(x)$ Proof: Taking $P(x,z-f(f(x)))$ for some $z > f(f(x))$ gives that $z > f(f(x)) \implies f(z) < f(x)$. In particular, taking $z = x$ gives that $f(f(x)) \ge x$ for all $x$. So we know $f^4(x) \ge f^2(x)$. But if we had $f^4(x) > f^2(x)$, then taking $z = f^4(x)$, we have that $f^2(x) \ge f^4(x)$, a contradiction. So we have $f^4(x) = f^2(x)$ for all $x$. $\square$ Let $S$ denote the set of all reals $s$ such that $f(f(s)) = s$, note that $f^2(x) \in S$ by Claim 1. $P(s,y)$ gives us $f(s) = f(s+y)(f(sf(y))+1)$ for $s \in S$. Call this $Q(s,y)$. Claim 2: $f(x) = \frac{c}{x}$ for all $x \in S$. Proof: Let $m,n$ be two elements of $S$, note that $f(m), f(n)$ are also in $S$. Dividing $Q(m,f(n))$ by $Q(n,f(m))$ and comparing it with dividing $Q(f(m), n)$ by $Q(f(n), m)$, we obtain $$\frac{f(m)}{f(n)} = \frac{f(m+f(n))}{f(n+f(m))} = \frac{n}{m}$$so the value $mf(m)$ is constant over $S$ and so there exists some $c$ such that $f(x) = \frac{c}{x}$ for all $x \in S$. $\square$ Claim 3: $f$ is injective. Proof: Suppose not, and $f(a) = f(b)$ with $a \neq b$. Then taking $y = a$ and $y = b$ in the previous equation, we get $f(s+a) = f(s+b)$. Comparing $P(x,a)$ and $P(x,b)$, we get $$f(f(f(x)) + a) + f(x+a)f(xf(a)) = f(f(f(x))+b) + f(x+b)f(xf(b))$$but by the previous paragraph, since $f(f(x)) \in S$, we must have that $f(f(f(x)) + a) = f(f(f(x))+b)$ so $f(x+a) = f(x+b)$ for all positive reals $x$. So this means $f$ is periodic with period $p = |a-b| > 0$ for all $x > \min(a,b)$. But then $P(x,x+np-f(f(x)))$ for sufficiently large $n$ and $x > a+b$ gives a contradiction, so $f$ is indeed injective. $\square$ Since $f(f(x)) = f(f(f(f(x))))$ and $f$ is injective, this means that $f(f(x)) = x$ for all $x$ so $S = \mathbb{R}^+$, since we had $f(x) = \frac{c}{x}$ for all $x \in S$, it follows that $f(x) = \frac{c}{x}$ for all $x \in \mathbb{R}^+$, so we are done. $\blacksquare$

AAAAAGH. This is about the same difficulty as That One USEMO FE, a little disappointed I didn't get this one in contest. edit: lolno I was just being bad The answer is all functions of the form $f(x)=\frac{c}{x}$ for constant $c$. Check they work. Note $f^2(x)\geq x$ as otherwise $y=x-f^2(x)$ gives $f(x)=f(x)+f(xf(y))f(x+y)>f(x)$, contradiction. If $f^2(a)>f^2(b)$ we claim $f(a)<f(b)$. Plugging in $x=b$ and $y=f^2(a)-f^2(b)$ gives \[f(b)=f^3(a)+f(xf(y))f(x+y)>f^3(a)\geq f(a)\]as desired. Then we claim $f^2(x)=f^4(x)$ for all $x$. If not then $f^2(x)<f^4(x)$, implying $f(x)>f^3(x)$ due to above, contradiction as $f(x)\leq f^3(x)$. Now we prove $f$ is injective. Let $f(a)=f(b).$ Then \[f^3(x)=f(f^4(x)+a)+f(f^2(x)f(a))f(f^2(x)+a)\]implies \[\frac{f^3(x)}{1+f(f^2(x)f(a))}=f(f^2(x)+a)\]Note LHS is just in terms of $x$ and $f(a)$, so $f(f^2(x)+a)=f(f^2(x)+b)$. Now $x,a$ gives \[f(x)=f(f^2(x)+a)+f(xf(a))f(x+a)\]so $f(x+a)=f(x+b)$. Let $b\geq a$ and $b-a=d$; then for all $x>a,$ we have $f(x)=f(x+kd)$ for integer $k$. This means that $f^2(x+kd)=f^2(x)\geq x+kd$ for any integer value of $k$. This is only possible if $d=0$; otherwise, $x+kd$ is unbounded while $f^2(x)$ is a constant. Thus $a=b$. Injectivity then gives $x=f^2(x)$. So substituting $x=1$ gives $\frac{f(1)}{y+1}=f(y+1),$ or $f(x)=\frac{f(1)}{x}$ for $x>1.$ Now we prove this for numbers less than $1$. Fix $x>1$ and then take some $y>1$ such that $xf(y)<1$, taking advantage of the fact that $f$ can be arbitrarily small. Then \[f(x)=f(x+y)(1+f(xf(y))\]\[\frac{f(1)}{x}=\frac{f(1)}{x+y}(1+f(xf(y))\]\[\frac{y}{x}=f(xf(y))=f(\frac{xf(1)}{y})\] Obviously $\frac{xf(1)}{y}$ can take on any real between $0$ and $1$. If we let $\frac{xf(1)}{y}=r$ then the equation can be rewritten as $\frac{f(1)}{r}=f(r)$ as desired.

dchenmathcounts wrote: AAAAAGH. This is about the same difficulty as That One USEMO FE, a little disappointed I didn't get this one in contest. The answer is all functions of the form $f(x)=\frac{c}{x}$ for constant $c$. Check they work. Note $f^2(x)\geq x$ as otherwise $y=x-f^2(x)$ gives $f(x)=f(x)+f(xf(y))f(x+y)>f(x)$, contradiction. Now we prove $f$ is injective. Let $f(a)=f(b).$ Then \[f(a)=f(f^2(a)+y)+f(af(y))f(a+y)\]implies \[f(a+y)=\frac{f(a)-f(f^2(a)+y)}{f(af(y))}.\]Note RHS is just in terms of $a$, so $f(a+y)=f(b+y)$. Hmph, why is $f(a+y)=f(b+y)$? How did you deal with $f(af(y)), f(bf(y))$? Also I think your argument for $f^{(3)}(x)=f(x)$ is wrong. You would only get $f^{(4)}(x)=f^{(2)}(x)$.

Argh, yaeh that is wrong. I mistakenly thought all of RHS was just wrapped around $f(a)$ when the denominator isn't. So injectivity probably looks like everyone else's solution, the $f^2(x),y$ substitution and letting $y$ be $a,b$ (since my solution gets $f=f^3$ independently of injectivity edit: apparently not, then f^3(x),y?). And then you get $f(f^2(x)+y)$ is a constant as long as $f(y)$ is (Will make edits to correct my solution) edit to edits: Yeah, I needed f^2(x),y before the f(x+a)=f(x+b) argument

The solution is pretty smooth! Let $P(x,y)$ denote the given assertion, and let $f^k(x)$ denote the $k^{\text{th}}$ composition of $f.$ Since the codomain of $f$ is $\mathbb{R}_{>0},$ from $P(x,y)$ it follows that $f(x)>f(f^2(x)+y)$ for all $x,y\in\mathbb{R}_{>0}.$ Let this property be $Q(x,y).$ Claim 1: For all $x\in\mathbb{R}_{>0}$ we have $f^4(x)=f^2(x).$ Proof: Firstly notice that if $f^2(x)<x$ for some $x,$ then $Q(x,f^2(x)-x)$ yields $f(x)>f(x),$ which is absurd. Thus, we can conclude that $f^2(x)\geq x$ for all $x\in\mathbb{R}_{>0}.$ In particular, this implies that $f^4(x)\geq f^2(x).$ Assuming that $f^4(x)\neq f^2(x),$ then $Q(x,f^4(x)-f^2(x))$ gives us $f^5(x)<f(x).$ This is yet again absurd, since we have $f(x)\leq f^3(x)\leq f^5(x)$ as per our previous observation. Thus, for all $x\in\mathbb{R}_{>0}$ we have $f^4(x)=f^2(x).$ $\blacksquare$ Claim 2: The function $f$ is injective. Proof: Assume, for the sake of contradiction, that $f(a)=f(b)$ for some $a\neq b.$ We say that a number $x$ is good if $x=f^2(y)$ for some $y.$ Claim 1 implies that $f^2(x)=x$ for all good numbers. By comparing $P(y,a)$ and $P(y,b)$ for some good $y$ we can deduce that $f(y+a)=f(y+b).$ However, by looking at $P(y,a)$ and $P(y,b)$ for any $y$ (not necessarily a good $y$ this time) we get that \begin{align*}f(y)&=f(f^2(y)+a)+f(yf(a))\cdot f(y+a) \\ &=f(f^2(y)+b)+f(yf(b))\cdot f(y+b)\end{align*}Notice that $f^2(y)$ is good, and thus $f(f^2(y)+a)=f(f^2(y)+b)$ (as per our argument on the previous line). Therefore, we can ultimately conclude that $f(yf(a))\cdot f(y+a)=f(yf(b))\cdot f(y+b)$ so $f(y+a)=f(y+b)$ for any $y\in\mathbb{R}_{>0}.$ The lattter is equivalent to the fact that $f$ is periodic (from a point on) with period $p:=|a-b|>0.$ This cannot hold, because by fixing $x$ (big enough for periodicity to apply) and taking $c\in\mathbb{N}$ big enough such that $c\cdot p+x-f^2(x)>0,$ we reach a contradiction because \[Q(x,x+c\cdot p-f^2(x))\implies f(x)>f(x+c\cdot p).\]Therefore, we cannot have $f(a)=f(b)$ with $a\neq b$ so $f$ is injective, as desired. $\blacksquare$ We finished with the hard part. By combining both claims we firstly infer that $f^2(x)=x$ for all $x\in\mathbb{R}_{>0}.$ Therefore, $P(1,y)$ yields $f(y+1)=f(1)/(y+1)$ so, in other words, $f(x)=f(1)/x$ for all $x>1.$ A couple of other plugins combined with the fact that $f^2(x)=x$ will ultimately imply that $f(x)=f(1)/x$ for all $x.$

We begin with some Claims ($f^k(x)$ denotes the $k-$th iteration of $f$): Claim 1: $y>f^2(x)$ implies $f(x)>f(y)$. Proof: Indeed, if $y>f^2(x)$ we may take $y \rightarrow y-f^2(x)$ in the given assertion, and so $f(x)=f(y)+f(xf(y-f^2(x)))f(x+y-f^2(x))>f(y)$, as desired $\blacksquare$ Claim 2: $f^2(x) \geq x$ for all $x>0$. Proof: Indeed, suppose that there existed a $x$ such that $x>f^2(x)$. By Claim 1, we would have $f(x)>f(x)$, a contradiction $\blacksquare$ Claim 3: $f^4(x)=f^2(x)$ for all $x>0$. Proof: Suppose that there existed a $x$ such that $f^4(x) \neq f^2(x)$. By Claim 2 we have $f^4(x)>f^2(x)$, and so by Claim 1 $f(x)>f^5(x)$. However, by Claim 2, $f(x)>f^5(x) \geq f^3(x) \geq f(x),$ a contradiction $\blacksquare$ Claim 4: $f(x)<\frac{f(1)}{x-1}$ for all $x>1$. Proof: Indeed, take $x=1$ and $y \rightarrow x-1$ in the given, and show $f(1)>f(f(x-1))f(x) \geq (x-1)f(x)$, which gives the desired result $\blacksquare$ Claim 5: $f$ is injective. Proof: Suppose there existed $a>b$ such that $f(a)=f(b)$. Then, take $x \rightarrow f^2(x)$ in the given, and so by using Claim 3 we have, $f^3(x)=f(f^2(x)+y)+f(f^2(x)f(y))f(f^2(x)+y),$ and now if we take $y=a$ and $y=b$ and comparing, we have $f(f^2(x)+a)=f(f^2(x)+b)$ for all $x$. Hence, by taking $y=a$ and $y=b$ in the given and comparing we get $f(x+a)=f(x+b)$. Thus, $f(x)=f(x+a-b)$ for all $x>a$, and subsequently $f(x)=f(x+k(a-b))$ for all $x>a$ and positive integers $k$. Now, fix $x$ such that $x>\max \{1,a \}$, and so $f(x+k(a-b))<\frac{f(1)}{x-1+k(a-b)}$, hence $f(x)<\frac{f(1)}{x-1+k(a-b)}$, which is a contradiction for $k$ pretty large. Thus, $f$ must be injective $\blacksquare$ To the problem, by Claims 3 and 5 we have $f^2(x)=x$, hence the given rewrites as $f(x)=f(x+y)(f(xf(y))+1)$. Take $x=1$ here, and so $f(x)=\frac{f(1)}{x}$ for all $x>1$. Now take a $x>\max \{1, \frac{1}{a} \}$ and $y=\frac{xa}{k}$ with $k \leq 1$. Then, $y=\frac{xa}{k}>\frac{1}{k} \geq 1$, and so we have $\frac{a}{x}=\frac{a}{x+y}(f(x \cdot \frac{k}{x})+1),$ hence $f(k)=\frac{a}{k}$ for all $k \leq 1$. Thus, we have $f(x)=\frac{f(1)}{x}$ for all $x$, which works.

Excellent problem! We begin by proving some trivial claims. Claim 1. \(f(f(x))\geq x\). Proof. Assume not. Then, \(P(x,x-f(f(x)))\) gives us a contradiction. $\blacksquare$ Claim 2. \(f^4=f^2\). Proof. I claim that if \(x>f(f(y))\), then \(f(y)>f(x)\). This is simply \(P(y,x-f(f(y)))\). This implies that if \(f(f(x))>f(f(y))\), then \(f(y)>f(x)\) by claim 1. Now, if \(f^4>f^2\), then \(f\le f^3<f\), a contradiction. And from claim 1, we must have \(f^4\ge f^2\), so I conclude that \(f^4=f^2\). $\blacksquare$ Claim 3. \(xf(x)\) is bounded. Proof. First of all, note that \(P(f^2(x), y)\) and claim 2 gives us that \[\frac{f^3(x)}{f(f^2(x)+y)}=1+f(f^2(x)f(y))\]Therefore, \[\frac{f^3(x)}{f(f^2(x)+f(y))}=1+f(f^2(x)f^2(y))=\frac{f^3(y)}{f(f^2(y)+f(x))}\]And so \[\frac{f^3(x)}{f(f^2(x)+f(y))}=\frac{f^3(y)}{f(f^2(y)+f(x))}\]Now replace \(x\) with \(f^2(x)\) and \(y\) with \(f(y)\). We see that \[\frac{f^5(x)}{f(f^4(x)+f^2(y))}=\frac{f^4(y)}{f(f^3(y)+f^3(x))}\]and as \(f^4=f^2, f^5=f^3\) we see that \[\frac{f^3(x)}{f(f^3(x)+f^3(y))}=\frac{f^2(y)}{f(f^2(x)+f^2(y))}\]Switching \(x\) and \(y\) and comparing, we see that \(f^2(x)f^3(x)\) is constant. Let this constant be \(c\). Now, \[c=f^2(x)\cdot f^3(x)\geq xf(x)\]and we're done. $\blacksquare$ Claim 4. \(f\) is injective. Proof. I claim that \(f\) is either injective or periodic. Indeed, let \(f(a)=f(b)\). Let \(\alpha\) be such that \(f^2(\alpha)=\alpha\). Then, \(P(\alpha,a)\) and \(P(\alpha,b)\) give us that \(f(a+\alpha)=f(b+\alpha)\). Now, note that \(\alpha\) can be \(f^2(x)\) for all \(x\), from claim 2. So, we have \(f(f^2(x)+a)=f(f^2(x)+b)\). Now, compare \(P(x,a)\) and \(P(x,b)\) to get \(f(x+a)=f(x+b)\) for all \(x\in\mathbb{R}^+\) and therefore \(f\) is periodic. Now, let the period be \(t\). Notice that \(P(x+t,y)\) and \(P(x,y)\) give us that \(f(x)=f(x+tf(y))\). The thing here is, \(tf(x)\) can be as small and as large as possible. Because from claim \(3\), choose \(x\) really large, this must imply \(f(x)\) really small and from claim 1, choosing \(x\) really large implies that \(f^2(x)\) is really large. We will use this as an advantage. So, we see that \[f(x+u)=f(x+v)\]where \(\lvert u-v\rvert\) is as large as possible. Now, let \(x_0\) be a fixed positive real so that \(v>f(x_0)>u\), note \(u\) is super small and \(v\) is super large. Then, replacing \(x\) with \(f(x_0)-u\) gives us that \[f(f(x_0))=f(f(x_0)+v-u)\]and remember \(f(f(x_0))>x_0\) and \(f(f(x_0)+v-u)<\frac{c}{f(x_0)+v-u}\). Therefore, \[x_0<f(f(x_0))=f(f(x_0)+v-u)<\frac{c}{f(x_0)+v-u}\]implying that \[x_0f(x_0)+x_0(v-u)<c\]But note that we can keep \(u\) the same and make \(v\) as big as we can! From the inequality, we MUST need \(x_0(v-u)<c\) for all possible values for \(v\), but if we make \(v\) tending to infinity, which we of course can, we get a contradiction. This forces \(f\) injective. $\blacksquare$ Now for the final blow, we have that \(f\) is an involution from claim \(1\). And as we got \(f^2\cdot f^3=c\), we see that \(xf(x)=c\), implying that \(f(x)=\frac{c}{x}\) for all positive reals \(x\), which is indeed, a solution. PS: I just realized how stupid I was, we have \(f(x+kt)=f(x)\) and let \(k\) be a super large positive integer then we are easily done on the injectivity argument. Moreover, we dont even need my claim 3, because once we get \(f\) is injective, then \(f\) is involutive, so \[f(x)=f(x+y)(1+f(xf(y)))\]plug \(x=1\) and so \(f(x)=\frac{c}{x}\) for all \(x>1\). Now let \(y>1\) and \(x>1\). \(x<\frac{y}{c}\). Note all \(t<1\) can be written as \(\frac{cx}{y}\). Then just do \(P(x,y)\) and finish for positive reals at most \(1\). Oops, a small thing, for the injectivity finish, we do require \(P(1,x-1)\) and then you dont need my claim 3, (else you do need lol) to get \(f(x)<\frac{f(1)}{x-1}\), my bad. I posted the above solution because that is what I found. Looking back at it, it does not feel too much like an AMO P3, but since I took ~2 hours, I'd rather not comment more on this.

USJL wrote: Also, the entire motivation behind this FE is...... Galois theory????? It's a long story and "Galois theory" is no longer involved in this version but yeah maybe I'll talk about it later. The backstory is typed up in the section 3.5 of the document here: https://artofproblemsolving.com/community/c6h2784600p24469107

megarnie wrote: Claim 1: If $f$ is involution, we are done. Proof: Suppose $f(f(x))=x$ for all $x$. $P(1,x): f(1)=f(1+x)+xf(x+1)\implies f(x+1)=\frac{f(1)}{x+1}$, as desired. $\blacksquare$ When I did it, I had to show more than this, since this only gives it for $x>1$.

jasperE3 wrote: megarnie wrote: Claim 1: If $f$ is involution, we are done. Proof: Suppose $f(f(x))=x$ for all $x$. $P(1,x): f(1)=f(1+x)+xf(x+1)\implies f(x+1)=\frac{f(1)}{x+1}$, as desired. $\blacksquare$ When I did it, I had to show more than this, since this only gives it for $x>1$. thanks. fixed.

Let $f^k(x)$ denote $f$ iterated $k$ times, and let $P(x,y)$ be the given assertion. Claim 1:: If $f(x)\le f(y)$, then $y\le f^2(x)$ (similarly if $y>f^2(x)$ then $f(x)>f(y)$) Both of the statements follow by letting $y>f^2(x)$ and taking: $P(x,y-f^2(x))\Rightarrow f(x)>f(y)$ Claim 2: $f^4(x)=f^2(x)$ Suppose $f^2(u)<u$ for some $u$. Then: $P(u,u-f^2(u))\Rightarrow f(uf(u-f^2(u)))f(2u-f^2(u))=0$ which is impossible, so $f^2(x)\ge x$ for all $x$. Since $f^5(x)\ge f^3(x)\ge f(x)$, we have $f(x)\le f^5(x)$. Then, $f^4(x)\le f^2(x)$ by Claim 1. From $f^2(x)\ge x$, we have $f^4(x)=f^2(x)$. Claim 3: $f$ is injective Suppose there are $a<b$ such that $f(a)=f(b)$. $P(f(f(x)),a),P(f(f(x)),y)\Rightarrow f(f^2(x)+a)=f(f^2(x)+b)$ (using Claim 2) $P(x,a),P(x,b)\Rightarrow f(x+a)=f(x+b)$ By induction, $f(2a)=f(2a+n(b-a))$ for $n\in\mathbb N$. Take $n$ large enough so that $n>\left|\frac{f^2(2a)-2a}{b-a}\right|+1000$, then $2a+n(b-a)>f^2(2a)$, so by Claim 1, $f(2a)>f(2a+n(b-a))$ which is absurd. Claim 4: $\boxed{f(x)=\frac ax}$ for $a\in\mathbb R^+$ Let $a=f(1)$. From $f^4(x)=f^2(x)$ we have $f^2(x)=x$, so $P(x,y)$ becomes: $$f(x)=(1+f(xf(y)))f(x+y).$$Now $P(1,x)\Rightarrow f(x+1)=\frac a{x+1}$ so $f(x)=\frac ax$ for $x>1$. Now let $y\le1$, and let $x>\max\left\{\frac1y,1-f(y)\right\}+100$. Then: $P(x,f(y))\Rightarrow f(x)=(1+f(xy))f(x+f(y))\Rightarrow \frac ax=(1+\frac a{xy})\cdot \frac a{x+f(y)}\Rightarrow f(y)=\frac ay$ so the claim is proven. $\frac ax$ is a solution for any positive $a$.

How much MOHS would you rate this one? Personally I would rate it 30 MOHS, although I believe 25 isn't that bad either

ZETA_in_olympiad wrote: How much MOHS would you rate this one? Personally I would rate it 30 MOHS, although I believe 25 isn't that bad either I'd say 30, as this was >>> 2016 USAMO 4

GoodMorning wrote: ZETA_in_olympiad wrote: How much MOHS would you rate this one? Personally I would rate it 30 MOHS, although I believe 25 isn't that bad either I'd say 30, as this was >>> 2016 USAMO 4 The two problems are different with different solutions, could you explain what you mean?

ZETA_in_olympiad wrote: GoodMorning wrote: ZETA_in_olympiad wrote: How much MOHS would you rate this one? Personally I would rate it 30 MOHS, although I believe 25 isn't that bad either I'd say 30, as this was >>> 2016 USAMO 4 The two problems are different with different solutions, could you explain what you mean? 16 USAMO/4 was a very standard real FE containing no unstandard tricks and which you can make nontrivial progress from trivial substitutions. Whereas for this one I found the jump to partial involution—and only being able to prove injectivity from that point on to be not as motivated.

GoodMorning wrote: 16 USAMO/4 was a very standard real FE containing no unstandard tricks and which you can make nontrivial progress from trivial substitutions. Whereas for this one I found the jump to partial involution—and only being able to prove injectivity from that point on to be not as motivated. P1s/P4s are typically standard, and this is P3 which is ought to be tough nevertheless

ZETA_in_olympiad wrote: GoodMorning wrote: 16 USAMO/4 was a very standard real FE containing no unstandard tricks and which you can make nontrivial progress from trivial substitutions. Whereas for this one I found the jump to partial involution—and only being able to prove injectivity from that point on to be not as motivated. P1s/P4s are typically standard, and this is P3 which is ought to be tough nevertheless he never brought up the difficulty relative to its placement

Let $P(x,y)$ be the assertion and let $f^n(k)$ be the $n$ iterations of $f.$ If $k>f^2(x),$ then $P(x,k-f^2(x))\implies f(x)>f(k).$ From this we have $f^2(x)\geq x\implies f^4(x)\geq f^2(x).$ But comparing with $k\mapsto f^4(x)$ we get $f^4(x)=f^2(x).$ Lemma: $f$ is an injection. Proof. Assume $\exists u>v: f(u)=f(v).$ Then comparing $P(f^2(x),u)$ and $P(f^2(x),v)$ yields $f(f^2(x)+u)=f(f^2(x)+v).$ Using this compare $P(x,u)$ and $P(x,v)$ to get $f(x+u)=f(x+v)$ but this forces $f$ to periodic, which is easily seen to be false. $\blacksquare$ $P(1,x-1)\implies f(x)=f(1)/x=c/x~\forall x>1.$ But it's not hard to conclude that it holds for all $x\in \mathbb{R}_{>0}.$

Not gonna lie, the question was really hard for me and I needed a hint to start, solved anyways. Claim 1. $f^2(x) \ge x$. Assume otherwise, then $P(x,x-f^2(x))$ gives $f(x)=f(x)+C$ where $C >0$, hence contraction. Putting $x= f^2(x)$, $f^4(x) \ge f^2(x) \ge x$ Claim 2. $f^4(x)=f^2(x)$. If our claim is not true, then $f^4(x)>f^2(x) \ge x$ and $P(x,f^4(x)-f^2(x))$ gives $$ f(x) = f^5(x) + C$$where $C>0$, which means $f(x)>f^5(x)$. But it is wrong since using Claim 1. we have $f(x) \le f^3(x) \le f^5(x)$, hence $f^2(x)=f^4(x)$ Claim 3. Function is injective Suppose $f(a)=f(b)$ where $b>a$ . Observing $P(f^2(x),a),P(f^2(x),b)$ we have $$f(f^2(x)+a)=f(f^2(x)+b)$$. Using the original equation, we have $f(x+a)=f(x+b)$. Since $x$ is arbitrary, we can make $b$ arbitrarly bigger such that $x+b > f^2(x+a)$, which implies ( from inequlity $f(x) > f(f^2(x)+y)$) $f(x+b)<f(x+a)$, hence contraction, $a=b$\ Since $f^4(x)=f^2(x)$, by injectivity, $f^2(x)=x$, which implies surjectivity.Then $$f(x)= f(x+y) ( 1 + f(xf(y)))$$Since plug in $x=1$ $$f(1) = f(1+y)(1+y)$$Let $x=1+y$, then $f(x)= \frac{f(1)}{x}$, which gives answer as $f(x) = \frac{c}{x}$ for constant $c$, which indeed satisfises the f.e

Let $P(x, y)$ denote the assertion. Note that $P(x, x - f^2(x))$ gives $f(x) = f(x) + \text{something}$, impossible, meaning that $x - f^2(x) \leq 0 \implies f^2(x) \geq x$ for all $x > 0$. Next, we prove $f$ injective. Suppose there is $0 < a < b$ for which $f(a) = f(b)$. Then, comparing $P(f^2(x), a)$ and $P(f^2(x), b)$ gives\[f(f^2(x) + a) = f(f^2(x) + b).\]Now, comparing $P(x, a)$ and $P(x, b)$ yields $f(x + a) = f(x + b)$, which is amazing because we can write $f(x + a) = f(x + C(b - a))$ for sufficiently large $C$. Now, we can write $f(x + a) > f(f(x + a)^2 + y)$ and make $y$ sufficiently large as well to make $f^2(x + a) + y = x + C(b - a)$ to force $f(x + a) > f(x + b)$, a contradiction. Hence, $a = b$, and $f$ injective. Now, we prove $f^4(x) = f^2(x)$. Else, $f^4(x) > f^2(x)$, which is what we assume. $P(x, f^4(x) - f^2(x))$ yields $f(x) > f^5(x)$, impossible since $f^5(x) \geq f^3(x) \geq f(x)$. Hence, $f^4(x) = f^2(x)$ and from injectivity $f$ is an involution with $f^2(x) = x$, and is thus bijective. Now, the FE becomes $f(x) = f(x + y)(1 + xf(y))$. Now, $x = 1$ yields $f(y + 1) = \tfrac{f(1)}{y + 1}$. This proves $f(x) = \tfrac Cx$ for $x > 1$. For specific $\epsilon < 1$, we can take $P(\epsilon, y)$ for large $y$ to show $f(\epsilon) = \tfrac {C}{\epsilon}$, proving that our final solution set is indeed just $f(x) = \tfrac Cx$.

Here is a rewrite of my solution in PM#3 that doesn't use injectivity. The answer is $f(x)\equiv \frac ax$ which clearly works. Let $P(x,y)$ denote the assertion in the question. Define $f^k(x) = f(f^{k-1}(x))$. Claim 1 $x\le f^2(x)$. Proof: Assume not, then $P(x,x-f^2(x))$ gives $f(x)=f(x)$+smth nonzero, contradiction. Corollary 1 $f^3(x)\ge f(x)$ by applying claim 1 on $f(x)$ (as $f(\mathbb{R}^+) \subseteq \mathbb{R}^+$) Claim 2: If $f^2(x)>f^2(y)$ then $f(x)<f(y)$. Proof: $P(y,f^2(x)-f^2(y))$ gives $f(y)>f^3(x)\ge f(x)$. Claim 3: $f^2(x)=f^4(x)$ for all $x\in \mathbb{R}^+$ Proof: Assume not. Then $f^2(x)<f^4(x)$ and $f(x)<f^3(x)$. Let $A=f(f(x))$, then we have $f(x)<f(A)$ and $f(f(x))<f(f(A))$, which contradicts claim 2. Claim 4: $f(x)<\frac{f(1)}{x-1}$ Proof: $P(1,x-1)$. Claim 5: $f(x)= f^3(x)$ for all $x\ge 1$ Proof: Let $B=f^2(x)$. We are given $f^2(x)=f^2(B)$ by claim 3. Assume $f(x)<f(B)$ since $f(x)\le f^3(x)$ by corollary 1. This also means $1\le x<B.$ Then $P(x,y), P(B,y)$ yields$$f(x)=f(f(f(x))+y)+f(x+y)f(xf(y))$$$$f(B)=f(f(f(B))+y)+f(B+y)f(Bf(y))$$ Let $t\to\infty$. Plugging in $y=f(t)$ and subtracting the second equation by the first give $f(B)-f(x)< f(B+f(t))f(Bf(f(t))) <\frac{f(1)}{B-1} \cdot \frac{f(1)}{Bf(f(t))-1} \le \frac{f(1)}{B-1} \cdot \frac{f(1)}{Bt-1}$ which tends to 0 as $t$ tends to infinity. Claim 6: $f( f^2(x)+f(y)) = f(x+f(y))$ for all $y\ge 1$. Proof: $P(f(x),y): f(f(x)) = f( f(x)+y ) ( 1 + f(f(x)f(y)))$ $P(f(x), f(f(y))): f(f(x)) = f( f(x)+f(f(y))) (1+f(f(x)f(y)))$ Thus, both quantities are equal to $\frac{f(f(x))}{1+f(f(x)f(y))}$ Claim 7: $f( xf(f(y))) = f( f(f(x)) \cdot f(f(y)))$ for all $x\ge 1$ $P(x,f(y))$: $f(x)=f( f(f(x))+f(y)) + f(x+f(y))f(xf(f(y)))$ $P(f(f(x)),f(y))$: $f(x)= f( f(f(x))+f(y)) + f( f(f(x))+f(y)) f( f(f(x)) f(f(y)))$ Since $f(x+f(y))=f(f(f(x))+f(y))$ it follows that $f(xf(f(y))) = f( f(f(x)) f(f(y)))$ for all $x\ge 1$ Finally, note $Im(f^3)\subseteq Im(f^2)\subseteq Im(f)=Im(f^3)$ (this is over $\mathbb{R}^+$) so they are all the same. In particular, $f( xf(y)) = f( f(f(x)) \cdot f(y))$ for all $x\ge 1$ Now, say $x\ge 1$. Letting $B=f(f(x))$. $P(B,y)-P(x,y)$ give $f(Bf(y))f(B+y)=f(xf(y))f(x+y)$. We just proved $f(Bf(y))=f(xf(y))$. Therefore, $f(B+y)=f(x+y)$. Setting $y=f(f(z))-x$ gives $B=x$, which proves $f(x)=\frac ax$ for $x\ge 1$. The rest is easy.

Denote the assertion with $P(x, y)$. We have that if $a > f(f(x))$, $f(a) < f(x)$. Equivalently, if $f(a) \ge f(x)$ then $a \ge f(f(x))$ As such, $f(f(x)) \ge x$. Claim: $f(f(x))$ has a fixed point. Proof. Note that $f(x) \le f^{3}(x) \le f^{5}(x)$. Then, if $f^{4}(x) > f(f(x))$ it follows that $f^{5}(x) < f(x)$, contradiction. Thus, $f(f(x)) = f^{4}(x)$ and $f^{2}(x)$ is a fixed point. $\blacksquare$ Claim: $f$ is strictly decreasing on $[c, \infty)$ for some $c$. Furthermore, $c \le f(f(x))$ for all $x$. Proof. Let $c$ be the minimal fixed point of $f \circ f$. Then, by $P(c, x)$ it follows that \[ f(c) = f(c + y)(1 + f(cf(y))) \]so $f(c) > f(c + y)$. $\blacksquare$ Claim: $f$ is injective and thus an involution. Proof. Suppose that $f(a) = f(b)$ for $a \ne b$. Then, by $P(f(f(x)), a)$ and $P(f(f(x)), b)$ \[ f(f(f(x)) + a) = f(f(f(x)) + b) \]However, $f$ is strictly decreasing on this domain, contradiction. Since $f^{4}(x) = f(f(x))$ it follows that $f(f(x)) = x$. $\blacksquare$ $P(x, y)$ thus simplifies as \[ f(x) = f(x + y)(1 + f(xf(y))). \]Thus, by $P(1, x)$ \[ \frac{f(1)}{f(1 + x)} = (1 + x) \]so $f(x) = \frac{f(1)}{x}$ for $x \ge 1$. Since $f$ is a involution this holds for all $x$. It can be seen that for all $f(x) = \frac{c}{x}$, $f$ is a involution and that \[ \frac{c}{x} = \frac{c}{x + y} + \frac{c}{x \cdot \frac{c}{y}} \cdot \frac{c}{x + y} \]works.

this problem repaired my marriage The answer is $f(x)=C/x$ only, which clearly work. First note that $$f(x)>f(f^2(x)+y),\qquad (\heartsuit),$$so $f^2(x) \geq x$, otherwise we can plug in $y=x-f^2(x)$ and find a contradiction. Let $f(1)=c$; by plugging in $x=1$, we find that $$c>f(f(y))f(1+y) \geq yf(y+1) \implies f(x)<\frac{1}{x-1}~\forall x>1.$$In particular, $\lim_{x \to \infty} f(x)=0$. I claim that $f^4(x)=f^2(x)$ for all $x$. Suppose otherwise, so $f^4(x)>f^2(x)$. On the other hand, we also have $f^5(x)\geq f^3(x) \geq f(x)$ for all $x$. Then by $(\heartsuit)$, we get a contradiction by setting $y=f^4(x)-f^2(x)$, because now $f(x)>f^5(x)$. Suppose that for some $a$, $f^3(a)>f(a)$. Then by replacing $x$ with $f^2(x)$ in the original FE and using the fact that $f^4(x)=f^2(x)$, we obtain $$f^3(x)=f(f^2(x)+y)(1+f(f(x)^2f(y)))~\forall x.$$Then by comparing $y=f(a)$ and $y=f^3(a)$ we find that $f(f^2(x)+f(a))=f(f^2(x)+f^3(a))$. Then by comparing $y=f(a)$ and $y=f^3(a)$ in the original FE we find that $f(x+f(a))=f(x+f^3(a))$ for all $x$. By iterating this it follows that $f(x+f(a))=f(x+f^3(a))=f(x+(2f^3(a)-f(a)))=\cdots$, and since $f^3(a)>f(a)$ we find an unbounded "chain" of positive reals which produce the same output when $f$ is applied, but this contradicts $\lim_{x\to\infty}f(x)=0$. Thus $f^3(x)=f(x)$ for all $x$. Then by repeating the exact same argument by supposing that $f^2(a)>a$ for some $a$, we find that $f^2(x)=x$ for all $x$, i.e. $x$ is an involution. With this new information, rewrite the FE as $$f(x)=f(x+y)(1+f(xf(y))).$$Replace $x$ with $f(x)$ to find that $$\frac{x}{f(f(x)+y)}=1+f(f(x)f(y)).$$Swapping $x$ and $y$ then implies that $$\frac{x}{y}=\frac{f(f(x)+y)}{f(f(y)+x)}.$$But then simultaneously replacing $x$ and $y$ with $f(x)$ and $f(y)$ respectively yields $$\frac{f(x)}{f(y)}=\frac{f(f(y)+x)}{f(f(x)+y)}=\frac{y}{x},$$hence $xf(x)$ is constant, i.e. $f(x)=C/x$ for some $x$, which is the desired result. $\blacksquare$

The answer is $f(x) = \boxed{\tfrac{c}{x}}$ for $c>0$. This is shown to work, so we will prove it is the only solution. Denote the given assertion as $P(x,y)$ and at any point in this proof, $f^n(x)$ denotes $n$ iterations of $f$. Consider the following assertion: \[Q(a,b): \ f(a) \ge f(b) \implies f(f(b)) \ge a. \] To see that this is true, we will consider the contrapositive. If $a > f(f(b))$, then $P(b,a-f(f(b)))$ yields \[f(b) = f(a) + \text{blah} > f(a),\] which finishes the proof. Claim 1: $f^2(x) = f^4(x)$ Proof: Notice that $Q(a,a)$ yields \[f^2(a) \ge a.\] Then, plugging in $a = f(x)$ and $a=f^3(x)$ gives \[f^5(x) \ge f^3(x) \ge f(x).\] This means $Q(f^4(x), x)$ holds; in other words, \[f^2(x) \ge f^4(x).\] But, we know that $f^4(x) \ge f^2(x)$ by plugging in $a = f(x)$. Therefore, $f^2(x) = f^4(x)$ as desired. $\square$ Claim 2: $f$ is injective Proof: Suppose that $f(a) = f(b)$ for two reals $a>b$. Comparing $P(f^2(x),a)$ and $P(f^2(x),b)$ yields \[f(f^4(x)+a) + f(f^2(x)f(a)) f(f^2(x)+a) = f(f^4(x)+b) + f(f^2(x)f(b)) f(f^2(x)+b)\]\[\implies f(f^2(x)+a) \left[1 + f(f^2(x)f(a)) \right] = f(f^2(x)+b) \left[1 + f(f^2(x)f(b)) \right].\] This implies that $f(f^2(x)+a) = f(f^2(x)+b)$. Thus, $f$ is periodic. However, taking an arbitrarily large value of $k$ such that $f(x) = f(x+k)$ and $f^2(x) < x+k$ leads to a contradiction due to $Q$. $\square$ From injectivity, Claim 1 becomes $f(f(x)) = x$. Plugging this into $P$ yields \[f(x) = f(x+y)+f(xf(y))f(x+y) = f(x+y) \left(1+f(xf(y)) \right).\] Substituting in $x=1$ yields \[f(1) = f(y+1) (1+f(f(y)))\]\[ \implies f(y+1) = \frac{f(1)}{y+1},\] which means $f(x) = \tfrac{c}{x}$, as desired.

Not very difficult after proving $f$ is injective.

We claim $f(x)=\frac cx$ for constant $c$. First notice $f(f(x))\ge x$, as otherwise $P(x,x-f(f(x)))$ gives a contradiction. Next, if $f(f(x))<f(y)$, then $P(x,f(y)-f(f(x)))$ implies $f(x)>f(f(y))\ge y$. Now we claim $f(f(x))=f(f(f(f(x))))$. If $f(f(x))<f(f(f(f(x))))$, then $f(x)>f(f(f(x)))$, but we have $f(f(f(x)))\ge f(x)$. Next by taking $P(f(f(x)),f(y))$ and $P(f(f(x)),f(f(f(y))))$ gives $f(f(f(x))+f(y))=f(f(f(x))+f(f(f(y))))$. Now $P(x,f(y))$ and $P(x,f(f(f(y))))$ give $f(x+f(f(f(y))))=f(x+f(y))$. Thus either $f(f(f(y)))=f(y)$ for all $y$, or $f$ is eventually periodic with period $f(f(f(y)))-f(y)=p>0$. However this is impossible since it would imply $f(f(x))=f(f(x+kp))\ge x+kp$ for $x$ large enough and all positive integers $k$. Now take $P(x,f(f(y)))$. We get \begin{align*}f(x)&=f(f(f(x))+f(f(y)))+f(xf(y))f(x+f(f(y)))\\&=f(f(f(x))+f(f(y)))+f(xf(y))f(y)-f(xf(y))f(yf(x))f(x+y).\end{align*}Thus by symmetry we have \[\frac{f(x)}{f(y)}=\frac{1+f(yf(x))}{1+f(xf(y))}.\] Now if $f(a)=f(b)$ then setting $x=a,x=b$ we get $f(af(y))=f(bf(y))$ for all $y$. Now $P(a,y),P(b,y)$ give $f(a+y)=f(b+y)$, so $a=b$ or $f$ is eventually periodic, which is impossible. Thus $f$ is injective, so $f(f(x))=f(f(f(f(x))))$ implies $x=f(f(x))$. Finally, setting $x,y=f(x),f(y)$ in $\frac{f(x)}{f(y)}=\frac{1+f(yf(x))}{1+f(xf(y))}$ gives $xf(x)=yf(y)$, so $f(x)=\frac cx$ for some constant $c$.

The answer is $f(x)=\tfrac{c}{x}$ for all $c\in \mathbb{R}_{>0}$, which clearly works. Claim: $f(f(x))\ge x$. Assume $f(f(x_0))<x_0$ then $P(x_0,x_0-f(f(x_0)))$ gives \[f(x_0f(y))f(x_0+y)=0\]which is a contradiction because the range of $f$ does not include $0$. Claim: $f^4(x)=f^2(x)$. We already know that $f^4(x)\ge f^2(x)$. Suppose $f^4(x_0)>f^2(x_0)$ for some $x_0$ then $P(x_0,f^4(x_0)-f^2(x_0))$ gives \[f(x_0)=f^5(x_0)+f(x_0f(y))f(x_0+y)>f^5(x_0)\]but $f^5(x_0)\ge f^3(x_0)\ge f(x_0)$, so that's a contradiction. Claim: $f$ is injective. Taking $P(f^2(x),y)$ gives \[f^3(x)=f(f^2(x)+y)(1+f(f^2(x)f(y)))\]Suppose $f(y_1)=f(y_2)$ with $y_1\neq y_2$ then $f(f^2(x)+y_1)=f(f^2(x)+y_2)$ for all $x$. Now take regularly $P(x,y_1)$ and $P(x,y_2)$ gives \[f(x+y_1)=\frac{f(x)-f(f^2(x)+y_1)}{f(xf(y_1))}=\frac{f(x)-f(f^2(x)+y_2)}{f(xf(y_2))}=f(x+y_2)\]which implies $f$ is eventually periodic. Let $k$ be its fundamental period, then for all $x$, $P(x,k)$ gives \[f(x)=f^3(x)+f(xf(y))f(x)\]but $f^3(x)\ge f(x)$, so this is impossible. Since $f^4(x)=f^2(x)$, we have $f^2(x)=x$, so our equation simplifies to \[f(x)=f(x+y)(1+f(xf(y)))\]Let $Q(x,y)$ denote the above assertion. Comparing $Q(f(x),y)$ and $Q(f(y),x)$ gives \[\frac{x}{f(f(x)+y)}=\frac{y}{f(f(y)+x)}\]while comparing $Q(x,f(y))$ and $Q(y,f(x))$ gives \[\frac{f(x)}{f(x+f(y))}=\frac{f(y)}{f(y+f(x))}\]These two together imply that $xf(x)=yf(y)$ which implies $f(x)=c/x$ for some $c$. We are done.