Is it true that if $x_n$ has at least 2 distinct terms, then $\liminf_{n\to\infty} (x_1 + \dots + x_n - n\sqrt[n]{x_1\dots x_n}) > 0$?

Question

Is it true that for every positive sequence $x_n$ with at least 2 distinct terms, $\liminf_{n\to\infty} (x_1 + \dots + x_n - n\sqrt[n]{x_1\dots x_n}) > 0$?

The equality condition of the AM-GM inequality gives $x_1 + \dots + x_n - n\sqrt[n]{x_1\dots x_n} > 0$ for all $n$ large enough but this is not enough to decide. I tried to find a counterexample to no avail either; I investigated $x_n = \frac{1}{n}$ and $x_n = 1 + \frac{1}{2^n}$ but I could not obtain a contradiction.

If the statement is true or false can we explain intuitively/heuristically why we expect it to be such?

Any comments and help are welcome.

Answer 1

Assume for instance that $x_1\neq x_2$. Let $\delta=x_1+x_2-2\sqrt{x_1x_2}>0$. Then $x_1+x_2+\ldots+x_n =\delta+\sqrt{x_1x_2}+\sqrt{x_1x_2}+x_3+\ldots+x_n \geq \delta+n\sqrt[n]{x_1\ldots x_n}$, QED.

Answer 2

Yes it is true. According to Bound for difference between arithmetic and geometric mean the following general lower bound holds: $$\sum_{k=1}^n x_k-n\left(\prod_{k=1}^n x_k\right)^{1/n}\geq \sum_{k=1}^n (\sqrt{x_k}-\mu_n)^2$$ where $\mu_n=\frac{1}{n}\sum_{k=1}^n \sqrt{x_k}$.

Then the RHS is greater than a positive number (which does not depend on $n$) as soon as we have at least $2$ distinct terms, say $x_i$ and $x_j$, $$\sum_{k=1}^n (\sqrt{x_k}-\mu_n)^2\geq (\sqrt{x_i}-\mu_n)^2+(\sqrt{x_j}-\mu_n)^2\geq \frac{(\sqrt{x_i}-\sqrt{x_j})^2}{2}>0$$ where $n\geq i,j$