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}

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} This is the part that I stuck. Maybe I choose the wrong way. What do you think? Any thoughts would be helpful. Thanks

$$(a_1a_2 \cdots a_n)^{1/n}= \left(\frac{1}{n!}\prod_{j=1}^nja_j\right)^{1/n} \leqslant \frac{1}{n(n!)^{1/n}}\sum_{j=1}^n j a_j$$
From Stirling's approximation we have $$(n!)^{-1/n} \leqslant e/(n+1)$$ and, thus,
$$\sum_{n=1}^k(a_1a_2 \cdots a_n)^{1/n }\leqslant e \sum_{n=1}^k\frac{1}{n(n+1)}\sum_{j=1}^nj a_j = e \sum_{n=1}^k\frac{1}{n(n+1)}\sum_{j=1}^kj a_j \mathbf{1}_{j \leqslant n}\\ = e \sum_{j=1}^kj a_j\sum_{n=1}^k\frac{1}{n(n+1)}\mathbf{1}_{j \leqslant n}= e \sum_{j=1}^kj a_j\sum_{n=j}^k\frac{1}{n(n+1)} \leqslant e \sum_{j=1}^kj a_j\sum_{n=j}^\infty\frac{1}{n(n+1)}\\ = e \sum_{j=1}^kja_j \cdot \frac{1}{j}= e\sum_{j=1}^ka_j = e \sum_{n=1}^k a_n$$