Suppose $f:\mathbf{N} \to [0,1]$ satisfies $$f(x)f(y) + f(x+y)\leq 1\qquad(1)$$ for all $x,y$. Let $$d_n = \frac{1}{n} \sum_{x=0}^{n-1} f(x).$$ It is easy to prove that $$\limsup d_n \leq 1/\varphi,$$ where $1/\varphi$ is the positive solution to $t^2+t=1$. (Start with $(1)$, take an average and $\limsup$ in $x$, then in $y$.)

What's interesting is that it does not seem nearly so easy to bound $d_n$, without the $\limsup$, independently of $f$. Is it true that $$d_n \leq 1/\varphi + o(1),$$ where the $o(1)$ is independent of $f$?

This does seem to be true numerically. In fact it may even be that $f \equiv 1/\varphi$ is the unique maximizer of $d_n$ for each $n$.