This question is a cross-post from MSE, cause I didn't get any answer there. I hope it is well suited for MO:


Let $(x\_{n})\_{n\ge 1}$ be an increasing sequence of positive integers and $\displaystyle{\overline{x}\_{n}:=\dfrac{1}{n}\sum\_{i=1}^{n}x\_{i}}$. Suppose furthermore that $\forall\varepsilon\gt 0, \ \ n.\overline{x}\_{n}\ll\_{\varepsilon}n^{1+\varepsilon}$. Does it entail that $x\_{n}\ll \overline{x}\_{n}^{2}$?  

Thanks in advance.


Edit: I think I must clear up my idea here. Writing $m\_{n}:=\dfrac{x\_{1}+x\_{n}}{2}$ and using the inequality of arithmetic and geometric means, one gets $x\_{1}.x\_{n}\le m\_{n}^{2}$. My idea is that, under the growth condition above, $m\_{n}$ is roughly of the same order of magnitude as $\overline{x}\_{n}$.



is $x_{n}\ll \overline{x}_{n}^{2}$?
No. Let $(y\_k)$ be a rapidly increasing sequence of positive integers, and for $y\_k\leq n<y\_{k+1}$ put $x\_n:=y\_k$. Note that $(x\_n)$ is an increasing sequence of positive integers satisfying $x\_n\leq n$. In particular, $\overline{x}\_{n}<n/2$. On the other hand, for $n:=y\_k$ we have $x\_n=y\_k$, while $\overline{x}\_{n}<1+y\_{k-1}$. Hence $x\_{n}\ll \overline{x}\_{n}^{2}$ would imply that $y\_k\ll y\_{k-1}^2$ which is not always true. For a concrete counterexample take $y\_k:=2^{3^k}$.


