I know for a fact that
$$\frac{1}{n}∑_{k=1}^n \Big(x_k^2 + 2\sqrt{3}(1 - \tfrac{2k-1}{n}) x_k + 1 \Big)>0 \qquad\text{if $x_1<x_2<…<x_n$}$$
should hold because I derived this sum as the squared Wasserstein-2 distance between an empirical distribution and the mean-0, variance-1 uniform distribution.
As a sanity check, I was trying to directly prove that it is positive, however, it appears non-trivial to do so since it is not non-negative term-by-term.
Each term in your sum is a quadratic polynomial of the form $x^2+2bx+c$, and as such, must have a minimum. The minimum value is $-b^2+c$.
Now let's get a lower bound by substituting each term with its minimum value. Your expression can't be less than $$\frac1n\sum_{k=1}^n \Big(-3(1 - \tfrac{2k-1}{n})^2+1\Big)= \frac1{n^3}\sum_{k=1}^n \Big(-3(n-2k+1)^2+n^2\Big)=\dots$$ (note that $n-2k+1$ changes in symmetric bounds, hence any sum of it times a constant is zero) $$\dots=\frac1{n^3}\sum_{k=1}^n \Big(6k(n-2k+1)+n^2\Big)=\frac1{n^3}\Big(6(n+1)\sum_{k=1}^nk -12\sum_{k=1}^nk^2+n^3\Big)=\dots$$ (skip the details) $$\dots=\frac1{n^3}\Big(n(- n^2+1) + n^3\Big)=\frac1{n^2}$$ which is obviously positive. Q.e.d.
As for the condition $x_1<\dots<x_n$ which we never used, we may just as well go and donate it to the poor.