Suppose we're given a filtered algebra $A$ over a field $k$ with filtration $F_{\bullet}A$ over the subspaces of $A$: $$\{0\}\subseteq F_{0}A\subseteq\cdots\subseteq F_{i}A\subseteq \cdots\subseteq A,$$ and suppose that $$\mathrm{gr}_{\bullet}^{F} A := \bigoplus_{i\in \mathbb{N}_0} \mathrm{gr}_{i}^{F}A$$ is the associated graded algebra of $A$, where $\mathrm{gr}_{i}^{F}A:=F_{i}A/F_{i-1}A$ and $\mathrm{gr}_0^FA=F_0A$.
Question: If $\mathrm{gr}_{\bullet}^{F}A$ is commutative, does it follow that $F_{i+j}A\subseteq F_{i}A\cdot F_{j}A$ for all $i, j \in \mathbb{N}_0$?
I couldn't think of a counterexample and the obvious ways that one would try to prove this don't lead me to anything useful.
Nope. To keep things really simple let's take $A$ itself to be commutative and graded (with the filtration where $F_n(A)$ is sums of elements of degree at most $n$, so the associated graded is just $A$ again): specifically, take
$$A = k[x_1, x_2]/(x_1^2 = x_2^2 = 0)$$
where $\deg x_i = i$. (It doesn't really matter whether we impose $x_2^2 = 0$ or not, the point is just to have an element of degree $2$ that isn't a sum of products of elements of degree $1$.)