By Kedlaya's result, we have: $$LHS\le 2021\sqrt[2021]{a_1\cdot \frac{a_1+a_2}{2}\cdot ...\cdot\frac{a_1+a_2+...+a_{2021}}{2021}}\le \frac{2021}{\sqrt[2021]{2021!}}$$so we need $\frac{2021}{\sqrt[2021]{2021!}}\le 3$, or, more generally, $\frac{n}{\sqrt[n]{n!}}\le 3$ for any natural $n$. This is obvious for $n=1$, and for the inductive step we need $(\frac{n+1}{3})^{n+1}\le (n+1)(\frac{n}{3})^n$, i.e. $(1+\frac{1}{n})^n\le 3$, which is obvious since $(1+\frac{1}{n})^n\le e< 3$.

This is a weak version of the classical Carleman Inequality \[\sum_{n=1}^{\infty} \sqrt[n]{a_1a_2\dots a_n} \le e \sum_{n=1}^{\infty} a_n\]which is easy to prove by AM-GM.

Also similar to this https://artofproblemsolving.com/community/c6h1273221p6663760.