A problem appeared in USAMO $2022$. Solve the functional equation $f: (0,+\infty) \to (0, +\infty):$
$$f(x) = f(f(f(x))+y) + f(xf(y))\times f(x+y)$$
I tried putting $x=1$,$y=1$ alternatively, using differentiation is also not proving to be helpful so what other methods can be applied? Thanks in advance.
We claim that the only solution is $\forall x, f(x) = c/x$ for any $c > 0$. This clearly satisfy the conditions. Now, we shall prove that this is the only solution.
Let $f_{(n)}$ denote $f(x)$ iterated $n$ times and $P(x, y)$ denote the given statement.
Claim: Let $Q(a, b)$ denote the statement $f (a) ≥ f (b) \implies f (f (b)) ≥ a$. Then, $Q$ must be true.
Proof: Suppose by contradiction that $a > f (f (b))$. Then, consider $P (b, a − f (f (b)))$ to get a contradiction.
Claim: Any function $f: \mathbb{R}^+ \to \mathbb{R}^+$ obeying statement $Q$ satisfies $f_{(2)}(x) = f_{(4)}(x)$.
Proof: From $Q(t, t)$ we get $f_{(2)}(t) ≥ t$ for all $t > 0$. So this already implies $f_{(4)}(x) ≥ f_{(2)}(x)$ by choosing $t = f_{(2)}(x)$. It also gives $f (x) ≤ f_{(3)}(x) ≤f_{(5)}(x)$ by choosing $t = f (x)$, $t = f_{(3)}(x)$. Then $Q(f_{(4)}(x), x)$ is valid and gives $f_{(2)}(x) ≥ f_{(4)}(x)$, as needed.
Claim: $f(x)$ is injective.
Proof: Suppose $f (u) = f (v)$ for some $u > v$. From $Q(u, v)$ and $Q(v, u)$ we have $f_{(2)}(v) ≥ u$ and $f_{(2)}(u) ≥ v$. Note that for all $x > 0$ we have statements:
$$P (f_{(2)}(x), u) \implies f_{(3)}(x) = f (x + u) + f (xf (u))f (x + u) = (1 + f (xf (u)))f (x + u) \\ P (f_{(2)}(x), v) \implies f_{(3)}(x) = f (x + v) + f (xf (v))f (x + v) = (1 + f (xf (v)))f (x + v)$$ It follows that $f (x + u) = f (x + v)$ for all $x > 0$. This means that $f$ is periodic with period $T = u − v > 0$. However, this is incompatible with $Q$, because we would have $Q(1 + nT, 1)$ for all positive integers $n$, which is obviously absurd.
Since $f$ is injective, we obtain that $f_{(2)}(x) = x$. Thus $P (x, y)$ now becomes the statement $f (x) = f (x + y)[1 + f (xf (y))]$
In particular $P (1, y) \implies f (1 + y) = \frac{f (1)}{1+y}$ so $f$ is determined on inputs greater than $1$. Finally, if $a, b > 1$ we get $P (a, b) \implies \frac{1}{a} = \frac{1}{a + b}[1 + f( \frac{a}{b} f (1))]$ which is enough to determine f on all inputs, by varying $(a, b)$.
Source: Evan Chen. Sorry, I typed this in a rush, so please comment if there are any errors.