Does there exist $\alpha>0, \beta\in (0,1)$ such that $\dfrac{\sum_{k=1}^n a_k}{n}\le \alpha (a_1\cdots a_n)^{1/n} + \beta \max_i(a_i)$ holds?

Let $$a_1\ge a_2\ge \cdots\ge a_n\ge 0$$ be given non-negative numbers. My question is the following:

Is there any $$\beta \in(0,1),\ \alpha>0$$, such that $$\dfrac{a_1+\cdots+a_n}{n}\le \alpha (a_1\cdots a_n)^{1/n}+\beta a_1?$$

This article derives inequalities of above sort, with $$\alpha=1$$ and $$\beta =1/n$$, but with $$\sum_{1\le i instead of $$a_1$$, multiplied with $$\beta$$. Also their method does not seem to be applicable to address my question. Is there any known result on this or is this an open question? Any ideas regarding how should one proceed? Thanks in advance.

• Let $\alpha=1/n,\beta=(n-1)/n$. Then $\alpha(a_1\dots a_n)^{1/n}\geq a_n,\beta a_1\geq (a_1+\dots+a_{n-1})/n$. – Wojowu Mar 11 at 19:18
• Oh right! That's such a simple observation. Thanks a lot! – Samrat Mukhopadhyay Mar 11 at 19:22
• @Wojowu: you probably mean $\alpha (a_1\ldots a_n)^{1/n} \ge a_n/n$. – Henri Mar 12 at 9:29
• @Henri Of course, silly typo. – Wojowu Mar 12 at 9:52