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?
> 
> 
> 


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**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


$$\sum\_{n\ge1}MG(a\_1,\dots,a\_n) \le\, e\, \sum\_{k\ge1} \bigg( \sum\_{n\ge k} \frac{1}{n(n+1)}\bigg) \, k a\_k =\, e\, \sum\_{k\ge1}\, a\_k \, .$$



Bounding the series of the geometric means of the terms of a given positive series
You can’t have $C< e$. Fix $N$, and define
$$a\_n=\begin{cases}\tfrac1n&n\le N\\\\\\\\0&\text{otherwise.}\end{cases}$$
Then
$$\sum\_na\_n=H\_N=\log N+O(1),$$
and
\begin{multline}\sum\_n\mathrm{MG}(a\_1,\dots,a\_n)=\sum\_{n=1}^N\frac1{\sqrt[n]{n!}}=\sum\_{n=1}^N\frac en\left(1+O\left(\frac{\log n}n\right)\right)\\\\=eH\_N+O(1)=\left(e+O\left(\frac1{\log N}\right)\right)H\_N.\end{multline}


