functional equation $f(xf(y)) + f(yf(x)) = 1 + f(x + y)$

Question

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.