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Functional equation $f(xf(y)) + f(yf(x)) = 1 + f(x + y) \space$ [B6 Putnam $2022$]

How do we find all smooth functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that$$f(xf(y)) + f(yf(x)) = 1 + f(x + y)$$for all $x, y > 0$?

By comparing degrees, the only polynomial solution is $f(x) = 1$. We can deduce from above that $\inf f \leq 1$, and $\sup f \geq 1$. But can we show that $f(x) = 1$ for all $x > 0$? Or is this not the solution?

This question is inspired by question B6 from the $2022$ Putnam Competition.


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