I stumbled upon the following inequality which seems to hold numerically, but which I can't fully prove. Let $x_1,....,x_n$ be a sequence of positive real numbers. Then $$\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}-\left(\prod_{i=1}^{n}x_{i}\right)^{1/n}\right)^{2}\leq\frac{n-1}{n}\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\left(\prod_{i=1}^{n}x_{i}^{2}\right)^{1/n}\right)$$
It is not too difficult to prove slightly weaker version with the dimension dependence $\frac{n-1}{n}$ term with 1 using a rescaling and Jenson type argument. By homogeoneity of both sides, we can assume without loss of generality that $\prod_{i=1}^{n}x_{i}=1$. Working from the LHS,
$$\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}-\left(\prod_{i=1}^{n}x_{i}\right)^{1/n}\right)^{2}=\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}-1\right)^{2}$$
$$=\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)^{2}+\left(1-\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)-\frac{1}{n}\sum_{i=1}^{n}x_{i}$$
We then have that the first term $\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)^{2}\leq\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}$ by Jenson, the second term $\left(1-\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)\leq0$ by AMGM, and the third term $\frac{1}{n}\sum_{i=1}^{n}x_{i}\geq1$ by AMGM as well, since we assumed that $\prod_{i=1}^{n}x_{i}=1$. Hence,
$$\left(\frac{1}{n}\sum_{i=1}^{n}x_{i}-1\right)^{2}\leq\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-1$$
$$=\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\left(\prod_{i=1}^{n}x_{i}^{2}\right)^{1/n}$$
One can extend this general proof idea to replace 2 with a general power $p>1$, but I cannot see how to bring in the n dependent term in front. It's also pretty easy to prove for $n=2$, but that is only easy because it's essentially minimizing one dimensional function (assume one of the numbers is 1, or only work with the ratio).
For $n=1$ it's obvious.
Let $n\geq2$, $\prod\limits_{i=1}^nx_i=constant$ and $\sum\limits_{i=1}^nx_i=constant$.
Thus, by the Vasc's EV Method $\sum\limits_{i=1}^nx_i^2$ gets a minimal value for equality case of $n-1$ variables.
Let $\prod\limits_{i=1}^nx_i=1$ and $x_2=...=x_n=x$.
Thus, $x_1=\frac{1}{x^{n-1}}$ and we need to prove that $f(x)\geq0,$ where $$f(x)=2n(n-1)x^{2n-1}-n(2n-1)x^{2n-2}-2(n-1)x^{n}+2nx^{n-1}+n-2.$$ Now, $$f'(x)=2n(n-1)x^{n-2}(n(2n-1)x^{n-1}-1)(x-1),$$ which says $x_{min}=1$ and since $f(0)\geq0$ and $f(1)=0,$ we are done.
About EV see here: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf (corollari 1.9, case 1)