Prove that $\sum_{n=1}^k (a_1\cdots a_n)^{1/n}< e \sum_{n=1}^k a_n, \forall a_1,\dots, a_k>0$ and $ k\in\mathbb{N}$ I tried to prove it by induction $\displaystyle\sum_{n=1}^k (a_1\cdots a_n)^{1/n}< e \displaystyle\sum_{n=1}^k a_n,\ \forall a_1,\dots, a_k>0$ and $ k\in\mathbb{N}$.
For $k=1$ the inequality holds. We suppose that the inequality holds for $k$, and we will show that it holds for $k+1$. Then
$$ \sum_{n=1}^{k+1} (a_1\cdots a_n)^{1/n}<e \sum_{n=1}^k a_n+ (a_1\cdots a_{k+1})^{1/(k+1)}. \tag{$\star$}$$
Also, by Jenssen's inequality (about $\ln x$), we have \begin{align*}(a_1\cdots a_{k+1})^{1/k+1}& =e^{\dfrac{\ln(a_1\cdots a_{k+1})}{k+1}}=e^{\dfrac{\ln(a_1)+\ldots + \ln(a_{k+1})}{k+1}}\\ & \leq e^{\displaystyle\ln \left(\dfrac{a_1+\ldots +a_{k+1}}{k+1}\right)}= \dfrac{a_1+\ldots +a_{k+1}}{k+1}=\dfrac{1}{k+1}\sum_{n=1}^{k+1} a_n. \end{align*} Thus, by $(\star)$ we have
$$\sum_{n=1}^{k+1} (a_1\cdots a_n)^{1/n}<e\displaystyle\sum_{n=1}^k a_n+\dfrac{1}{k+1}\sum_{n=1}^{k+1} a_n=\left(e+\frac{1}{k+1}\right) \sum_{n=1}^k a_n+\frac{1}{k+1}a_{k+1}.$$
This is the part that I stuck. Maybe I choose the wrong way.  What do you think? Any thoughts would be helpful. Thanks
 A: By the AM-GM inequality
$$(a_1a_2 \cdots a_n)^{1/n}= \left(\frac{1}{n!}\prod_{j=1}^nja_j\right)^{1/n} \leqslant \frac{1}{n(n!)^{1/n}}\sum_{j=1}^n j a_j$$
From Stirling's approximation we have $(n!)^{-1/n} \leqslant e/(n+1)$ and, thus,
$$\sum_{n=1}^k(a_1a_2 \cdots a_n)^{1/n }\leqslant e \sum_{n=1}^k\frac{1}{n(n+1)}\sum_{j=1}^nj a_j = e \sum_{n=1}^k\frac{1}{n(n+1)}\sum_{j=1}^kj a_j \mathbf{1}_{j \leqslant n}\\ = e \sum_{j=1}^kj a_j\sum_{n=1}^k\frac{1}{n(n+1)}\mathbf{1}_{j \leqslant n}= e \sum_{j=1}^kj a_j\sum_{n=j}^k\frac{1}{n(n+1)} \leqslant e \sum_{j=1}^kj a_j\sum_{n=j}^\infty\frac{1}{n(n+1)}\\ = e \sum_{j=1}^kja_j \cdot \frac{1}{j}= e\sum_{j=1}^ka_j = e \sum_{n=1}^k a_n$$
A: Just for the sake of completeness, I will prove RPLs statement that $(n!)^{-1/n}\leq e/(n+1)$. This holds if and only if:
$$-\frac{\sum_{j=1}^n \ln (j)}{n}\leq 1 -\ln(n+1) \Leftrightarrow$$
$$\sum_{j=1}^n \ln (j) \geq -n+n\ln(n+1) \Leftrightarrow$$
$$\sum_{j=1}^{n+1} \ln (j) \geq 1-(n+1)+(n+1)\ln(n+1)=\int_1^{n+1}\ln(x)dx  $$
This follows immediately by noticing that the left hand side could start at $j=2$ and is an upper Darboux sum, with the partition $P=\{1,2,...,n,n+1\}$.
