Let $ \{ a _ k \} _{k\in\mathbb{N} _ +} $ be a sequence of non-negative numbers, and let $MG(a_1,\dots,a_n)$ denote the geometric mean of the first $n$ terms. Then, the inequality $$ \sum _ {n\ge 1}MG(a_1,\dots,a_n) \le C\ \sum _ {n\ge 1} a _ n $$ holds, with $C=e$. This is quite elementary, although not obviously true (for instance, no analogous inequality could hold for the arithmetic means $MA(a_1,a_2,\dots,a_n)$, as the series on the LHS may then diverge even for a converging series on the RHS).
Questions: What is the name of the above inequality? Is $C=e$ the best constant for it? Is it attained?
$$*$$ edit. (Details on the above inequality). From the Arithmetic-Geometric means inequality $$MG(a_1,\dots,a_n)=MG(1a_1,2a_2,\dots,na_n)(n!)^{-1/n}\le MA(1a_1,2a_2,\dots,na_n)(n!)^{-1/n}\ .$$ Stirling formula in form of inequality, $n!\ge \sqrt{2\pi n}\ n^n e^{-n}$, written for $n+1$, implies $$(n!)^{-1/n} \le \frac{e}{n+1}$$ for all $n\ge1$. So
$$MG(a_1,\dots,a_n) \le \frac{e}{n(n+1)}\ \sum_{1\le k \le n} k a_k \ ,$$
whence
$$\sum_{n\ge1}MG(a_1,\dots,a_n) \le\ e\ \sum_{k\ge1} \bigg( \sum_{n\ge k} \frac{1}{n(n+1)}\bigg) \ k a_k =\ e\ \sum_{k\ge1}\ a_k \ .$$
