How to minimize this quadratic form?

Question

For real numbers $x_1,\dots, x_{2n}$, satisfying $\sum_{i=1}^{2n}x_i = 0$ and $\sum_{i=1}^{2n}x_i^2 = 1$, how do I show that $$ \sum_{1\le i < j\le n}(x_i-x_j)^2 + \sum_{n+1\le i < j\le 2n}(x_i-x_j)^2 + \sum_{i=1}^n\sum_{j=0}^{m-1} (x_i-x_{n+1+(i+j-1)})^2 \ge 2m$$ where the $(i+j-1)$ in the subscript is taken modulo $n$ and $m\le n$? A case of equality would be $x_i = \begin{cases}\frac{1}{\sqrt{2n}} & 1\le i\le n \\ -\frac{1}{\sqrt{2n}} & n+1\le i\le 2n\end{cases}$.

Answer 1

The problem most likely requires $m\le n$. Otherwise there are counterexamples.

It is more convenient to look at the complement of the big sum, namely $$\sum_{i,j}^{2n}(x_i-x_j)^2-\sum_{i=1}^n\sum_{j=0}^k(x_i-x_{f(i,j)})^2\ge2m$$ where $k=n-m$ and $f(j)$ refers to the complement indices to the ones specified by the asker, i.e. referring to the non-edges. Since the first sum equals $2n$, it is enough to prove $$\sum_{i=1}^n\sum_{j=0}^k(x_i-x_{f(j)})^2\le2k$$

The components $x_i$ ($1\le i\le n$) and $x_{f(i,j)}$ (effectively $x_{n+1},\ldots,x_{2n}$) are disjointed. Conceptually it is better to think of them as two distinct vectors, $x$ and $y$. The sum boils down to $$\|x-y\|^2+\|x-Ry\|^2+\cdots+\|x-R^ky\|^2$$ where $R$ is the cyclic-shift operator.

Claim: If $\sum_i(x_i+z_i)=0$ and $\sum_i(x_i^2+z_i^2)=1$ then $\|x-z\|^2\le2$.

Proof: Let $x_i=\alpha+x'_i$ where $\alpha=\sum_ix_i/n$; similarly $z_i=-\alpha+y'_i$. Then the condition $$1=\|x\|^2+\|z\|^2=2\alpha^2n+\|x'\|^2+\|z'\|^2$$ implies \begin{align*}\sum_i(x_i-z_i)^2&=\sum_i(2\alpha+x_i'-z_i')^2\\ &=4\alpha^2n+\|x'-z'\|^2\\ &\le4\alpha^2n+2\|x'\|^2+2\|z'\|^2\\ &=2 \end{align*}

Since $y$ and $R^jy$ satisfy the conditions of $z$ (for example $\|R^jy\|=\|y\|$), then $\|x-R^jy\|^2\le2$, so adding $k$ of them gives an upperbound of $2k$ as required.