Bulgarian MO 2021 & IMOC 2020 A2: https://artofproblemsolving.com/community/c6h2255027p17400924

That's unfortunate.

Who would expect that a problem that has appeared earlier was proposed in nationals, but anyways, here is my solution during the contest: Answer: $f(x) = 1$ for all $x \in \mathbb{R}^+$ or $f(x) = \frac{1}{x}$ for all $x \in \mathbb{R}^+$ Let $P(x, y)$ be the our assertion. Then $P(x, \frac{x-1}{x})$ yields that $f(\frac{x-1}{x} + f(x)) = 1$ for all $x > 1$. Fix a positive real $x$ and let $c = \frac{x-1}{x} + f(x)$. Now from $P(c, y)$, we deduce that $f(y + 1) = f(yc + 1)$ for all $y \in \mathbb{R}^+$. Claim: If $c \neq 1$, then $f(x)$ is a constant function. Proof. Let $x_1, x_2$ be positive reals such that $f(x_1) = f(x_2)$. Then $P(x_1, y)$ and $P(x_2, y)$ tell us that $f(x_1y + 1) = f(x_2y + 1)$ for all $y \in \mathbb{R}^+$, in other words, $f(x\frac{x_2}{x_1} + 1) = f(x + 1)$ for all $x \in \mathbb{R}^+$. Taking $x_2 = yc + 1$ and $x_1 = y + 1$, we see that $f(x\frac{yc+1}{y+1} + 1) = f(x + 1)$ for all $x, y \in \mathbb{R}^+$. Now it's not hard to check that $\frac{yc + 1}{y + 1}$ can attain any value that is between $1$ and $c$ when $y$ runs over $\mathbb{R}^+$ By proceeding induction on $k$, we can easily deduce that $f(xc^k + 1) = f(x + 1)$ for all $k \in \mathbb{Z}$. Now fix a positive reals $a, b$. We'll show that $f(a + 1) = f(b + 1)$. If $\frac{b}{a} = c^s$ for some $s \in \mathbb{Z}$, then certainly $f(a + 1) = f(b + 1)$. Thus, we may assume that there is an integer $k$ such that $\frac{b}{a}$ is between $c^{-k}$ and $c^{-k+1}$. Multiplying both sides by $c^k$, we see that $\frac{b}{a}c^k$ is between $1$ and $c$. Therefore, we can find a positive real $y$ such that $\frac{b}{a}c^k = \frac{yc + 1}{y + 1}$. $\implies f(a + 1) = f(a\frac{yc + 1}{y + 1} + 1) = f(bc^k + 1) = f(b + 1)$. Now just take $y > 1$ and plug into our equation, we deduce that $f(x) = 1$ for all $x \in \mathbb{R}^+$. $\blacksquare$ By claim, we may assume $c = 1$, so $\frac{x-1}{x} + f(x) = 1 \implies f(x) = \frac{1}{x}$ for all $x > 1$. Finally, taking $y > 1$ and plug into our equation, we see that $f(x) = \frac{1}{x}$. These answers satisfy the problem equation. This completes proof. $\blacksquare$

The answer is $f(x)\equiv 1$ and $f(x)=\frac{1}{x}$. Let $P(x,y)$ be the assertion in the hypothesis. For all real number $x_0>1$, $P(x_0,\frac{x_0-1}{x_0})$ gives $$f\left( \frac{x_0-1}{x_0} +f(x_0)\right)=1,\forall x_0>1.$$Then there exists real number $c$ such that $f(c)=1$. $P(c,y)$ gives $f(y+c)=f(1+cy),\forall y\in \mathbb{R^+}$. If $f$ is injective, then $y+c=1+yc,\forall y\in \mathbb{R^+}$, so $c=1$. This means that $f(c)=1\Leftrightarrow c=1$. Thus $$\frac{x-1}{x}+f(x)=1,\forall x>1,$$or $$f(x)=\frac{1}{x},\forall x>1.$$For all positive real number $x$, let $y$ be a positive real number such that $y+f(x)>1$ and $1+xy>0$, then from problem, we get $$f(x)\cdot \frac{1}{y+f(x)}=\frac{1}{1+xy},\forall x\in \mathbb{R^+}.$$Then $f(x)=\frac{1}{x},\forall x\in \mathbb{R^+}$. If $f$ is not injective, then there exists two real positive numbers $a,b$ such that $f(a)=f(b)$ and $a>b$. $P(a,y)$ and $P(b,y)$ give $$f(1+ay)=f(1+by),\forall y\in \mathbb{R^+}.$$Let $T=\frac{b}{a}<1$, then $f(y+1)=f(Ty+1),\forall y\in \mathbb{R^+}$. Then we easily to see that $$f(y+1)=f(T^ny+1),\forall y\in \mathbb{R^+},\forall n\in \mathbb{Z^+}.$$Now, we will proof that $f$is constant. Suppose that there exists two positive real number $u,v$ such that $u\neq v$ and $f(u)<f(v)$. $P(x,\frac{y}{x})$ gives $$f(x)f\left(\frac{y}{x}+f(x)\right)=f(1+y),\forall x,y\in \mathbb{R^+}.$$Let $Q(x,y)$ be the assertion in this hypothesis. $Q(u,y)$ gives $$f(u)f\left(\frac{y}{u}+f(u)\right)=f(1+y),\forall y\in \mathbb{R^+}.$$$Q(v,yT^n)$ gives $$f(v)f\left(\frac{yT^n}{v}+f(v)\right)=f(1+yT^n)=f(1+y),\forall y\in \mathbb{R^+},\forall n\in \mathbb{Z^+}.$$Since $T<1$, then $\lim_{n\to +\infty}T^n=0$, so there exists a positive integer $n_0$ such that $v-u\cdot T^{n_0}>0$. Now, put $y\to \frac{uv(f(v)-f(u))}{v-uT^{n_0}}$ and $n\to n_0$ in two above equations, we get $f(u)=f(v)$, this is a contradiction. Then $f(x)\equiv c$, with $c$ is a constant positive real number, put $f(x)\to c$ to problem, we get $c^2=c$ or $c=1$ (since $c\neq 0$). Thus $f(x)\equiv 1$.