One of the hardest FE problems I've done. Got this after a lot of hours after the test and some hints The answer is $f(x) = \frac{1}{x}$ which satisfies since \[ xf(x+y) + f(xf(y) + 1) = \frac{x}{x+y} + \frac{y}{x+y} = 1 = f(xf(x)) \]for all $x,y \in \mathbb{R}^+$. We now prove that there are no other functions that satisfy the functional equation. Let $P(x,y)$ be the assertion of $x$ and $y$ to the given functional equation. Claim 01. $f(x) < f(f(1))$ for all $x > 1$. Proof. $P(1,y)$ gives us $f(y+1) + f(f(y) + 1) = f(f(1)) \Rightarrow f(y+1) < f(f(1)) \ \forall y \in \mathbb{R}^+$, which proves our claim. Claim 02. $f$ is injective. Proof. Suppose otherwise, that there exists $a < b$ such that $f(a) = f(b)$. $P(x,a)$ and $P(x,b)$ gives us \[ f(x+a) = f(x+b) \ \forall x \in \mathbb{R}^+ \]Thus, $f$ is periodic for all $x \ge c$, a constant $c$ with period $b - a$. Fix $x_1, y_1 \in \mathbb{R}^+$, $x_1 > c$ and pick $n \in \mathbb{N}$ such that $(x_1 + np)f(x_1 + y_1) \ge f(f(1))$ and $(x_1 + np)f(x_1) > 1$. Thus, $P(x_1 + np, y)$ gives us \[ (x_1 + np)f(x_1 + y_1) + f((x_1 + np)f(y_1) + 1) = f((x_1 + np)f(x_1)) \]However, this forces $f((x_1 + np)f(x_1)) > (x_1 + np) f(x_1 + y_1) \ge f(f(1))$, and since $(x_1 + np)f(x_1) > 1$, from the previous claim, we get that \[ f((x_1 + np)f(x_1)) \le f(f(1)) \]a contradiction. Claim 03. $f$ is involutive. Proof. $P(1,f(y))$ and $P(1,y)$ gives us \[ f(f(y) + 1) + f(f(f(y)) + 1) = f(y + 1) + f(f(y) + 1) \Rightarrow f(y + 1) = f(f(f(y)) + 1) \Rightarrow f(f(y)) = y \ \forall y \in \mathbb{R}^+ \] Claim 04. $xf(x) \le 1$. Proof. If there exists $a \in \mathbb{R}^+$ such that $af(a) > 1$, then $P(x,f(y))$ gives us \[ xf(x + f(y)) + f(xy + 1) = f(xf(x)) \]Use the cancelling trick on $P(a,y)$ and let $ay + 1 = af(a)$, which is possible as $af(a) > 1$, we get a contradiction. Claim 05. $f(xf(x)) = 1$ for all $x \in \mathbb{R}^+$. Proof. First, we prove $f(xf(x)) \le 1$ for all $x \in \mathbb{R}^+$. Assume otherwise, that there exists $b \in \mathbb{R}^+$ such that $f(bf(b)) > 1$. Notice that $P(b,y)$ gives us \[ f(bf(b)) = bf(b+y) + f(bf(y) + 1) \overset{ f(bf(y) + 1) < f(f(1)) = 1}{\le} bf(b+y) + 1 \overset{f(b+y) \le \frac{1}{b+y}}{\le} \frac{b}{b+y} + 1 \]Now, we have $f(bf(b)) \le \lim_{y \to \infty} \frac{b}{b + y} + 1$, which forces $f(bf(b)) \le 1$, a contradiction. Now, assume that there exists $c \in \mathbb{R}^+$ such that $f(cf(c)) < 1$. Let $f(cf(c)) = 1 - \varepsilon$. Take a constant $\Delta$ such that for all $y \ge \Delta$, then $f(y+1) < \varepsilon$. Let $\delta > \Delta$. From $P(1,\delta)$, we have \[ f(f(\delta) + 1) > 1 - \varepsilon \]Remembering $f(x,f(y))$ gives us $f(xf(x)) = f(xy + 1) + xf(x+f(y)) > f(xy + 1)$, then this motivates $P \left( c, f \left( \frac{f(\delta)}{c} \right) \right)$ which gives us \[ 1 - \varepsilon = f(cf(c)) > f(f(\delta) + 1) > 1 - \varepsilon \]a contradiction. Therefore $f(xf(x))$ is a constant. Since $f$ is injective, then $xf(x)$ is a constant as well, which means $f(x) = \frac{c}{x}$ is the only possible solution. Checking back, \[ 1 = f(xf(x)) = xf(x+y) + f(xf(y) + 1) = \frac{cx}{x + y} + f \left( \frac{cx}{y} + 1 \right) = \frac{cx}{x + y} + \frac{cy}{cx + y} \]for all $x,y \in \mathbb{R}^+$. Take $x = y = 1$, gives us $\frac{c}{2} + \frac{c}{c + 1} = 1$, which means $c(c + 1) = 2$. This gives $c = -2$ or $c = 1$. But since $c \in \mathbb{R}^+$, this gives $c = 1$. Hence, the only solution is $\boxed{f(x) = \frac{1}{x}}$, done.

As proven above $f$ is involutive and $xf(x)\le 1$ . Set $c=f(1)$ :$f(c)=1$ we have $c\le 1$. We will need the following get the four equations from plugging in $(x,y),(y,x),(f(x),f(y)),(f(y),f(x))$ and compare them to get : $$(x+y)f(x+y)=(f(x)+f(y))f(f(x)+f(y)).....(2)$$set $(c,f(y))$ and get $$1=cf(c+f(y))+f(cy+1)\le cf(c+f(y))+\frac{1}{cy+1}$$$$f(c+f(y))\ge \frac{1}{c+1/y}$$and use $(2)$ to get that $(y+1)f(y+1)=(c+f(y))f(c+f(y))\ge \frac{c+f(y)}{c+1/y}$ in other words $\lim_{y\to \infty}yf(y)=1$. Next use relation (from setting $x=1$) $ f(y+1)+f(f(y)+1)=1 $ and plug these into $(2)$ to get that $$(y+f(y)+2)f(y+f(y)+2)=c$$use $(2)$ again to get that $$(f(y+2)+y)f(f(y+2)+y)=(y+2+f(y))f(y+2+f(y))=c$$I rewrite the latter two relations $$(f(y+2)+y+4)f(f(y+2)+y+4)=(f(y+2)+y)f(f(y+2)+y)=c$$Now plug in the equation $(f(y+2)+y,4)$ to finally get that $$\frac{c(f(y+2)+y)}{f(y+2)+y+4}+f((f(y+2)+y)f(4)+1)=1$$It is easy to see that if we let $y\to \infty$ we will get that $c=1$ (we could have done something similar in other ways too). After that set $y=1$ get that $f(xf(x))=(x+1)f(x+1)$ then $f$-this to get $xf(x)=(x+2)f(x+2)$ and $(x+2m)f(x+2m)=xf(x)$ for any integer $m$. This and the fact that $\lim_{y\to \infty}yf(y)=1$ give that $xf(x)=1$ for any $x$.

Let $P(x,y)$ denote the given proposition. Claim 1: $f(x)>1$ $\implies$ $x<1$. Proof: Assume the contrary, i.e., $f(x)>1$ and $x \geq 1$ for some $x$. Let $y=x(f(x)-1)>0$. Then $P(x,y)$ gives $$f(xf(x)) > xf(x+y) \geq f(xf(x))$$Contradiction! $\blacksquare$ In particular, by Claim 1, if $f(x)>x$ for some $x$, then we must have $x <1$; otherwise, we would have $f(x) >x \geq 1$, which contradicts Claim 1. Claim 2: $f$ is injective. Proof: Assume FTSOC that there exist $a>b$ such that $f(a)=f(b)$. Then comparing $P(x,a)$ and $P(x,b)$, we get $f(x+a)=f(x+b)$ for all $x$ $\implies$ $f$ is eventually periodic with period $p=a-b>0$, Now, $P(x,p)$ gives $$f(xf(x))>xf(x+p)=xf(x)$$$\implies$ $xf(x)<1$ for all $x$ by Claim 1. But then, $$f(x)=f(x+np) < \frac{1}{x+np}$$for all positive integers $n$, and we obtain a contradiction by taking $n \rightarrow +\infty$. $\blacksquare$ Claim 3: $f(f(y)) =y$ for all $y$. Proof: $P(1,y)$ gives $$f(y+1)+f(f(y)+1)=f(f(1))$$for all $y$. Putting $f(y)$ instead of $y$ in the above and using injectivity, we get $f(f(y))=y$. $\blacksquare$ Claim 3 in particular implies that $f$ is bijective. Claim 4: $f(x) \leq \frac{1}{x}$ for all $x$. Proof: Assume $xf(x)>1$ for some $x$. Then by surjectivity we can choose a $y$ such that $xf(y)+1=xf(x)$. But now $P(x,y)$ gives $f(xf(x))>f(xf(x))$, contradiction! $\blacksquare$ Let $S=\{f(x) \mid x>1 \}$. Let $m= \sup S$, which exists since $f(x)<1$ for all $x>1$ by Claim 1 and Claim 3. Note that $z\in S$ $\implies$ $f(z)>1$ by Claim 3. Claim 5: $f(xf(x)) \geq m$ for all $x$. Proof: For any $x>0$, and for any $z \in S$, there exists a $y>0$ such that $xf(y)+1=f(z)$ by surjectivity. Then $P(x,y)$ $\implies$ $f(xf(x)) > f(f(z))=z$. Since this holds for all $z \in S$, we must have $f(xf(x)) \geq m$ for all $x$. $\blacksquare$ Claim 6: $f(xf(x)) \leq m$ for all $x$. Proof: Fix an $x$. Then $P(x,y)$ gives $$f(xf(x))=xf(x+y)+f(xf(y)+1) \leq \frac{x}{x+y} +m$$for all $y$. Then taking $y \rightarrow \infty$ in the above, we get $f(xf(x)) \leq m$. $\blacksquare$ Claim 5 and Claim 6 give $f(xf(x))=m$ for all $x$. Therefore injectivity gives $f(x)=\frac{c}{x}$ for some constant $c$. Putting this in $P(x,y)$, we get that $c=1$. Therefore $$\boxed{f(x)=\frac{1}{x} \ \ \forall x \in \mathbb{R}^+}$$

AWSOME F.E.!! (my respect to the creators of this beauty) Solved with Rama1728 (Caling $P(x,y)$ assertion and blah blah blah) Claim 1: if $f(x)>1$ then $x<1$ Proof: Use $P(x,x(f(x)-1))$ and assume that $f(x)>1$, now if $x>1$ then $$f(xf(x(f(x)-1))+1)=(1-x)f(xf(x))<0 \text{contradiction!!}$$For proving the next claim we might need $f(x)>1$ then $x<1$ which means $f(x)>x$ then $x<1$. Claim 2: $f$ is injective. Proof: Assume that its not injective then there exists $a>b$ such that $f(a)=f(b)$ and by $P(x,a)-P(x,b)$ for $x>b$ $$f(x+a)=f(x+b) \implies f \; \text{periodic with period} \; a-b$$Now using $P(x,a-b)$ and Claim 1 we get $$f(xf(x))=xf(x)+f(xf(a-b)+1)>xf(x) \implies f(x)<\frac{1}{x} \implies f(x)<\frac{1}{x+(a-b)k} \implies f(x)<0 \; \text{contradiction!!}$$On the last step $k$ was a posititve integer and we just let $k \to \infty$ to get the contradiction. Claim 3: $f$ is an involution. Proof: Use $P(f(y),1)-P(y,1)$ $$f(y+1)+f(f(y)+1)=f(f(y)+1)+f(f(f(y))+1) \implies f(y+1)=f(f(f(y))+1) \implies f(f(y))=y \implies f \; \text{involution}$$Claim 4: $f(x) \le \frac{1}{x}$ Proof: Just use $P \left(x, f \left(\frac{xf(x)-1}{x} \right) \right)$ and assume that there exists $x$ such that $f(x)>\frac{1}{x}$ $$xf \left(x+f \left(\frac{xf(x)-1}{x} \right) \right)=0 \; \text{contradiction!!}$$Claim 5: $f(n)=\frac{1}{n}$ for every posititve integer $n$. Proof: Using Claim 4 we get $f(1) \le 1$ and also assume that there exists $x$ such that if $x<1$ then $f(x)<1$. and this will mean that if $f(x)<1$ then $x<1$ but note that using involution on Claim 1 this will mean that $1>x>1$ which its not possible. Hence if $x<1$ then $f(x)>1$ and using Claim 3 on this we get if $f(x)<1$ then $x>1$ so this with Claim 1 and Claim 3 forces that $f(1)=1$. Now by $P(n,1)$ and easy induction. $$(n+1)f(n+1)=f(nf(n)) \implies f(n)=\frac{1}{n}$$On the final part we will use this when $n \to \infty$ so this means $\lim_{x \to \infty} f(x)=0$ Final part: Using Claim 4 on the F.E. and letting $y \to \infty$ we get $$f(xf(x)) \le \lim_{y \to \infty} \frac{x}{x+y}+\frac{1}{xf(y)+1}=1$$But using Claim 4 anda variant of Claim 1 we get that $f(xf(x)) \ge 1$. Hence the unique solution of the F.E. is $f(x)=\frac{1}{x}$ thus we are done

MathLuis wrote: Proof: Assume that its not injective then there exists $a>b$ such that $f(a)=f(b)$ and by $P(x,a)-P(x,b)$ $$f(x+a)=f(x+b) \implies f \; \text{periodic with period} \; a-b$$Now using $P(x,a-b)$ and Claim 1 we get $$f(xf(x))=xf(x)+f(xf(a-b)+1)>xf(x) \implies f(x)<\frac{1}{x} \implies f(x)<\frac{1}{x+(a-b)k} \implies f(x)<0 \; \text{contradiction!!}$$ Excuse me if I am mistaken, but how exactly does $P(x,a-b)$ imply $f(xf(x))=xf(x)+f(xf(a-b)+1)$ for $x<b$?

Oops, i forgot to mention that $x>b$ (Sorry)

I think this solution is new. Let $P(x,y)$ be the assertion that \[xf(x+y)+f(xf(y)+1)=f(xf(x))\] Claim 1: If positive real numbers $x$ and $a$ satisfy $x>a$, then $f(x)<\frac{f(af(a))}{a}$. Proof: By $P(a,x-a)$, we have \[f(af(a)=af(x)+f(af(x-a)+1)>af(x),\]so $f(x)<\frac{f(af(a))}{a}$. If $f(a)>1$, then we can let $x=af(a)$ in the above claim to get $a<1$. Taking the contrapositive of this statement, we see that $f(a) \leq 1$ if $a \geq 1$. Claim 2: If positive real numbers $x$ and $a$ satisfy $x>a$, then $f(x) \geq \frac{f(af(a))-1}{a}$ Proof: Notice that since $xf(y)+1 \geq 1$, we have $f(xf(y)+1) \leq 1$. By $P(a,x-a)$, we have \[af(x)=f(af(a))-f(af(x-a)+1) \geq f(af(a))-1,\]so $f(x) \geq \frac{f(af(a))-1}{a}$ Claim 3: $\lim_{x \to \infty}f(x)=0$. Proof: We first show that $\lim_{x \to \infty}f(x)$ exists. Consider the intervals \[\mathcal{A}_n=\left[\frac{f(nf(n))-1}{n},\frac{f(nf(n))}{n}\right)\]for positive integers $n$. We know that $f(x)$ lies in $\mathcal{A}_n$ if $x>n$. We claim that there exists a real number $L$ in $\mathcal{A}_n$ for every positive integer $n$. Assume FTSOC that there doesn't exist such $L$. Then, there must be a positive integer $m$ such that $\mathcal{A}_1 \cap \mathcal{A}_2 \cap \cdots \cap \mathcal{A}_m$ is nonempty, but $\mathcal{A}_1 \cap \mathcal{A}_2 \cap \cdots \cap \mathcal{A}_{m+1}$ is empty. However, that means $f(x)$ cannot exist for any $x>m+1$, a contradiction. Thus, there exists such $L$. Now, notice that for any real number $x \geq 2$, $L$ and $f(x)$ both lie in the interval $\mathcal{A}_{\lfloor x \rfloor-1}$, so $|L-f(x)|<\frac{1}{\lfloor x \rfloor-1}$. Since this converges to $0$, $\lim_{x \to \infty}f(x)$ must converge to $L$. Assume FTSOC that $L \neq 0$. By $P(x,y)$ for arbitrarily large $x$ and $y$, we obtain $xL+f(Lx+1)=f(Lx)$, a contradiction since the LHS grows unbounded while the RHS converges to $L$. Thus, $L=0$. Claim 4: $f$ is injective. Proof: Assume FTSOC that there exists positive real numbers $a$ and $b$ such that $f(a)=f(b)$ and $a \neq b$. We compare $P(x,a)$ and $P(x,b)$ to get $f(x+a)=f(x+b)$, so $f$ is eventually periodic. However, that means $\lim_{x \to \infty}f(x)$ cannot converge to $0$, a contradiction. Now, take $P(a,y)$ for fixed $a$ and arbitrarily large $y$. Then, we have \[\lim_{x \to 1}f(x)=f(af(a)),\]so $f(af(a))$ is constant. Thus, $f(xf(x))=f(yf(y))$ for all real numbers $x$ and $y$, so $xf(x)=yf(y)$. Therefore, we have $f(x) \equiv \frac{c}{x}$, and it's easy to obtain that $f(x) \equiv \frac{1}{x}$ is the only solution.

This was hard. Denote the assertion by $P(x,y).$ We solve the problem in a number of steps. Step 1: We show that $f(x)>1 \implies x<1.$ Assume not. Then $P(x,xf(x)-x)$ readily yields contradiction. As a corollary note that $f(x)>x$ if $x>1.$ Step 2: We show that $f$ is one to one. If $f(u)=f(v).$ Then $P(x,u)$ and $P(x,v)$ imply $f$ is periodic with period $u-v.$ Now $P(x,u-v)$ gives $f(xf(x))>xf(x).$ This means $f(x)<1/(x+c(u-v))$ if $u-v\neq 0.$ Take large $c$ to get contradiction. Step 3: We show that $f$ is an involution. This is obtained when $P(x,f(y))$ is compared with $P(x,y).$ Step 4: We show that $xf(x)\leq 1.$ If not then take such a $z.$ Then $P(z,f((zf(z)-1)/z))$ readily yields contradiction. Step 5: We show that $f(xf(x))\leq 1.$ We have $f(xf(x))=f(xf(y)+1)+xf(x+y)<x/(x+y)+1.$ Taking large $y$ forces $f(xf(x))\leq 1.$ Step 6: We show that $f(xf(x))=1.$ Assume that we have some positive $k$ and $z$ such that $f(zf(z))=1-k.$ If $v$ is bounded then $P(1,v)$ implies $f(f(v)+1)=1-f(v+1)>1-k.$ We have contradiction from $P(z,f(f(v)/z)).$ Step 7: We show $f(x)=1/x.$ We have that $f(x)=c/x$ for some constant $c.$ Checking gives $c=1$ and thus $f(x)=1/x$ which can be seen to work.

Solved with Aayuu and TheProblemIsSolved As usual, let $P(x,y)$ denote the given assertion. Suppose $f(x) > 1$ for some $x$, then $xf(x) > x$ and so $P(x,xf(x)-x)$ gives that $(1-x)f(xf(x)) = f(xf(y)+1) > 0$ so $x < 1$. Suppose $f(a) = f(b)$ for distinct $a,b$ then comparing $P(x,a), P(x,b)$ we have that $f(x+a) = f(x+b)$ for all $x$, which means $f$ is eventually periodic with period $p = |a-b|$. Taking $P(x,p)$ gives that $f(xf(x)) > xf(x)$. If $xf(x) > 1$, then by the previous paragraph, we get a contradiction, so actually $f(x) \leqslant \frac{1}{x}$ for all big enough $x$. But $f(x) = f(x+kp)$ so $f(x) \leqslant \frac{1}{x+kp}$ for all $k$, impossible as $f$ is positive, so $f$ must be injective. Now, compare $P(1,y)$ and $P(1,f(y))$ to get that $f(f(y)) = y$ for all $y$, due to injectivity. Note that since $f$ is surjective, if $f(x) > \frac{1}{x}$, then we can choose $f(y)$ so that $xf(y) + 1 = xf(x)$, giving a contradiction. Therefore, we have that $f(x) \leqslant \frac{1}{x}$ for all $x$. Claim: $\lim_{y \rightarrow \infty} f(xf(y)+1)$ exists and equals $f(xf(x))$ Proof: Note that $$f(xf(x)) - \frac{x}{x+y} \leqslant f(xf(y)+1) < f(xf(x))$$using the fact that $0 < xf(x+y) \leqslant \frac{x}{x+y}$. So taking $y$ large enough makes the claim true. $\square$ Since $\lim_{y \rightarrow \infty} f(y) = 0$ $\left(0 < f(y) \leqslant \frac{1}{y} \right)$, we must have, by the above claim, that $f(xf(x)) = \lim_{y \rightarrow 1^{+}} f(y)$, which is constant. So $xf(x)$ must be constant, and checking gives that only $f(x) = \frac{1}{x}$ works, so done. $\blacksquare$

Denote the assertion with $P(x, y)$. Then, by $P(1, x)$ \[ f(1 + x) + f(f(x) + 1) = f(f(1)) \]so $1 + x \ne f(1)$ over all $x$ and thus $f(1) \le 1$. Claim: $f(x) \le \frac{c}{x}$ for sufficiently large $x$ and some $c$. Proof. We have that \[ f(x + y) < \frac{f(xf(x))}{x} \]so $f(f(1))$ bounds $f$ on $(1, \infty)$. As such, this means that \[ f(x + y) > \frac{f(xf(x)) - f(f(1))}{x}. \]Let $c = \max\{1, f(f(1))\}$. For each $x$, if $f(x) < \frac{1}{x}$ then this follows. Else, if $f(x) > \frac{1}{x}$ then \[ f(x) < \frac{f(xf(x))}{x} < \frac{f(f(1))}{x} \]$\blacksquare$ As such, \[ \lim_{x \to \infty} f(x) = 0. \] Claim: $f$ is injective. Proof. Suppose $f(a) = f(b)$ for $a \ne b$. Then, by $P(x, a)$ and $P(x, b)$ it follows that \[ f(x + a) = f(x + b) \]and thus $f$ is periodic. However, this contradicts the limit. $\blacksquare$ Taking $y \to \infty$ gives that \[ \lim_{y \to 0} f(1 + y) = f(xf(x)) \]where the limit exists since $f(xf(y) + 1) \le f(xf(x))$ holds. This is independent from $x$, so $xf(x)$ is constant and thus $f(x) = \frac{a}{x}$, of which only $a = 1$ works.

Let $P(x,y)$ be the assertion above. First thing is we know that for all $x > 1$ we have $f(x) \leq 1$ as by taking $x+y = xf(x)$, it shows us if $f(x) > 1$, we must have $x < 1$ to avoid a contradiction in $f(xf(y)+1) = (1-x)f(xf(x)) > 0$. Hence if $f(x) > 1, x > 1$, this would be a contradiction. Now let us assume $f$ is not injective. Then we get $f(x+a) = f(x+b)$ for some certain $a,b$. We must have at some point $f(x) \leq \frac{1}{x}$ for all $x \geq B$. Notice, if this is not true, we have $f(c) > \frac{1}{c}$ for some arbitrarily large $c$. Then $cf(c) > 1$, which means \[cf(c+y) + f(cf(y) + 1) < f(cf(c)) < 1\]and $f(c+y) < \frac{1}{c}$. But this is a contradiction as we have by the not-injective condition, a periodicity above some lower bound where we can assume $c$ is some arbitrarily large number, giving $f(c+p) = f(c) > \frac{1}{c}$. Thus now, notice even if $xf(x) \leq 1$ for every $x \geq B$, we have by the periodicity, taking a $x+Np$, we get that $(x+Np)f(x+Np) = (x+Np)f(x)$, which must increase arbitrarily as $N$ increases, thus breaking the upper bound of $1$. This must mean we have $f$ is injective. $P(1,y)$, \[f(y+1) + f(f(y)+1) = f(f(1))\]$y \mapsto f(y)$, \[f(f(y)+1) + f(f(f(y))+1) = f(f(1))\] \[f(y+1) = f(f(f(y))+1)\]\[y = f(f(y))\]which gives bijective and involutive. Now then since we have involutive, we know all values of larger than $1$ is reached by numbers smaller than $1$ and vice versa. Notice by forcing $xf(y)+1 = xf(x)$, we know $f(y) = f(x) - \frac{1}{x}$ can not be made, giving $f(x) \leq \frac{1}{x}$ for all $x$. Hence $\lim_{x \rightarrow \infty} f(x) = 0$. Hence taking $y \rightarrow \infty$, we get $C = f(xf(x))$, where we know a subsequential limit exists as $f(xf(y)+1)$ is bounded. Hence $xf(x) = c \rightarrow f(x) = \frac{c}{x}$, subbing to get $f \equiv \frac{1}{x}$.

I think this is different. The answer is $f(x)=1/x$ only, which works since both sides equal $1$. Let $P(x,y)$ denote the assertion. From $P(1,y)$, we have $f(y+1)+f(f(y)+1)=f(f(1))$, hence $f(y+1)<f(f(1))$ for all $y$, or equivalently $f(y)<f(f(1))$ for $y>1$. Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$ with $a \neq b$. By comparing $P(x,a)$ with $P(x,b)$ we find that $f(x+a)=f(x+b)$, hence $f$ is eventually periodic with period $|a-b|:=T$. Now by considering $P(kT,1)$ for some integer $k$ and sending $k \to \infty$, $kTf(kT+1)$ goes to infinity as well since $f(kT+1)$ is eventually constant. Thus $f(kTf(kT))$ is unbounded as $k \to \infty$, but $f(kT)$ is eventually constant as well, hence we will certainly have $kTf(kT)>1$ eventually. But this contradicts $f(y)<f(f(1))$. $\blacksquare$ Revisiting $P(1,y)$ and comparing it with $P(1,f(y))$ yields $f(y+1)=f(f(f(y))+1)$, hence by injectivity $f(f(y))=y$, i.e. $f$ is an involution. Thus we can rewrite the assertion (by replacing $y$ with $f(y)$) as $xf(x+f(y))+f(xy+1)=f(xf(x))$. This implies that $xf(x) \leq 1$ for all $x$, since otherwise we may pick $y$ with $xy+1=xf(x)$ and get a contradiction. Now suppose that we have $a,b$ with $f(af(a))>f(bf(b))$. By comparing $P(a,y)$ with $P(b,ay/b)$ and subtracting, it follows that $af(a+f(y))-b(b+f(ay/b))=f(af(a))-f(bf(b))>0$. On the other hand, since $f$ is an involution we can make $f(y)$ arbitrarily large, and since $xf(x) \leq 1$ for all $x$, it follows that $af(a+f(y))$ becomes arbitrarily small. But it must always be at least $f(af(a))-f(bf(b))$, which is a contradiction. Hence $f(xf(x))$ is constant, so by injectivity $xf(x)$ is as well and $f(x)=c/x$ for some $c$. It is straightforward to plug in and verify that only $c=1$ works. $\blacksquare$

One of the most known $\mathbb{R}_+$ FE. As usual, let $P(x;y)$ be the assertion. Let $f(f(1))=k$. $P(1;y) \Rightarrow f(y+1)+f(f(y)+1)=k$, in particular that $f(x)<k, \forall x>1$. First, I will prove that $f(xf(x)) \le k$. Assume it is not the case and we have some $x_1$, which does not fit. $P(x_1;y) $ Together with $f(x_1f(y)+1) < k$ results in $f$ being bounded below by something bigger than $0$, for large enough inputs. Now putting back large $x$ and $y$ we get that $xf(x+y) <f(xf(x))$ meaning that for large enough $x$, $f(xf(x))>k$, and thus resulting in $xf(x)<1$, which contradicts. Now I will show that $xf(x) \le k$, this is true as $xf(x+y) <f(xf(x)) \le k$, and fixing $x+y$ and varying $y$ to be very small we conclude. Let's check that $f$ is injective. Assume otherwise and let $f(y_1)=f(y_2)$ and $y_1-y_2=c; c>0$. $P(x;y_1)-P(x;y_2) \Rightarrow f(x)=f(x+c) \forall x \ge y_1$. However this means that $f(x)=f(x+nc) \le \frac{k}{x+nc}$ for natural $n$, which again can't be. $P(1;y)-P(1;f(y)) \Rightarrow f(f(y))=y, \; \forall y$, because of injectivity. Now assume that $f(xf(x))=k-a$, $a>0$ for some $x$ and $a$. Than $f(xf(y)+1) <k-a$, meaning $f(f(xf(y))+1)>a$, however as $f$ is surjective, and $f(x)\le \frac{k}{x}$, we conclude that it is impossible for $a$ to be positive. Now we know $f(xf(x))=k=f(f(1))$ thus by injectivity that $f(x)=\frac{f(1)}{x}$, and by substituting back we get $f(x)=\frac{1}{x}$, for any positive real $x$ which indeed is a solution.

The only solution is $\boxed{f(x) = \frac 1x }$, which works. Now we show it's the only solution. Let $P(x,y)$ be the given assertion. Claim 1: For $x \ge 1$, we have $f(x) \le 1$. Proof: Suppose for some $x \ge 1$ that $f(x) > 1$. $P(x, xf(x) - x)$ gives that $xf(xf(x)) <f(xf(x)) \implies x < 1$, absurd. $\square$ Now let $(1)$ denote the resulting inequality\[ f(xf(x)) \le xf(x+y) + 1\]Claim 2: $f$ is injective Proof: Suppose we had $f(a) \ne f(b)$ for some $a < b$ with $b - a = d$. $P(x,a)$ compared with $P(x,b)$ gives that $f(x+a) = f(x+b)$ for all $x > 0$, so $f(x) = f(x+d) \forall x \ge a$. Setting $y = d$ for $x \ge a$ gives that $f(xf(x)) = xf(x) + f(xf(d) + 1) > xf(x)$. This implies that $xf(x) < 1$, so $f(x) < \frac 1x \forall x \ge a$. Thus, $f(x) \to 0$ as $x \to \infty$. Now, for any fixed $x>0$, taking $y$ sufficiently large in $(1)$ (so that $f(x+y)$ is sufficiently small) gives that $f(xf(x)) \le 1$ for any positive real $x$. However, we also have $f(xf(x)) > xf(x+y)$, so $xf(x+y) < 1$. Since $f(x) = f(x+nd) \forall n \in \mathbb N$, replacing $x$ with $f(x + nd)$ gives that $(x+nd) f(x+y) = 1$. Taking $n$ arbitrarily large gives a contradiction. $\square$ Claim 3: $f$ is an involution Proof: $P(1,y)$ compared with $P(1, f(y))$ gives\[f(y+1) = f(f(f(y)) + 1)\implies y + 1 = f(f(y)) + 1 \implies f(f(y)) = y. \ \ \ \square\]Claim 4: As $x \to \infty$, $f(x) \to 0$. Proof: Suppose not and we had $f(x) > c $ for arbitrarily large $x$ (where $c>0$ is some constant). Now for any fixed $x$ so that $f(x) > c$, choose $y$ so that $f(x+y) > c$. We have $f(xf(x)) > cx$. Now choose $x$ large enough so that $cx > 1$. We get that $f(xf(x)) > 1 \implies xf(x) < 1$, so $c < f(x) < \frac 1x$, which means $cx < 1$, absurd. $\square$ Claim 5: For every constant $c < 1$, there exists $x$ with $x > 1$ and $f(x) > c$. Proof: $P(1,y): f(y+1) + f(f(y) + 1) = 1$. Choosing $y$ large enough so that $f(y+1) < 1 - c$ gives the desired result. $\square$ Claim 6: $xf(x) \le 1$ and $f(xf(x)) \ge 1$ for all positive reals $x$. Proof: $P(x, f(y)): xf(x + f(y)) + f(xy + 1) = f(xf(x))$, so $f(xf(x)) > f(xy + 1)$. Setting $y \to \frac yx$ gives\[ f(xf(x)) > f(y+1)\]If $f(xf(x)) < 1$, then choosing $y$ so that $f(y+1) > f(xf(x))$ (from claim 5) gives a contradiction. Therefore, $f(xf(x)) \ge 1$ for all positive reals $x$. It remains to show that $xf(x) \le 1$ for all $x > 0$. Suppose for some $x$ that $xf(x) > 1$. Since $f(xf(x)) \le 1$, we have $f(xf(x)) = 1$, so $xf(x) = f(1)$, absurd since $f(1) \le 1$. $\square$ We now have\[ f(xf(x)) \le \frac{x}{x+y} \cdot (x+y) f(x+y) + 1 \le 1 + \frac{x}{x + y}\]Setting $y$ arbitrarily large gives a contradiction if $f(xf(x)) > 1$, so $f(xf(x)) = 1$ for all positive reals $x$. Since $f$ is injective, we have $xf(x) = f(1)$ for all $x$, so $f(x) = \frac{f(1)}{x}$. Claim 7: $f(1) = 1$. Proof: Note that if not, then $f(1) < 1$. We then have for all $x > 1$ that $f(x) = \frac{f(1)}{x} < f(1)$. However, by claim 5 we can choose $x$ with $x > 1$ and $f(x) > f(1)$, a contradiction. $\square$ Thus $f(x) = \frac 1x$ must hold.