Look at the expression 
$$
f(x\_1,x\_2,x\_3) = x\_1^2+x\_2^2+x\_3^2+x\_1x\_2+x\_2x\_3+x\_3x\_1. 
$$
The numbers $x\_1,x\_2,x\_3$ are non-negative, and I assume that $x\_1+x\_2+x\_3=3$. This is a sum of squares and "cyclic correlations" of consecutive variables. Then you can check that $f$ is minimized for the values $x\_1=x\_2=x\_3=1$.


Now look at
$$
g(x\_1,x\_2,x\_3,x\_4) = x\_1^2+x\_2^2+x\_3^2+x\_4^2+x\_1x\_2+x\_2x\_3+x\_3x\_4+x\_4x\_1,
$$
under the assumption that $x\_1+x\_2+x\_3+x\_4=4$. Again, this is minimized by $x\_1 = x\_2 = x\_3 = x\_4=1$.


A similar thing happens if I add "second-order cyclic correlations": Let
$$
h(x\_1,x\_2,x\_3,x\_4) = x\_1^2+x\_2^2+x\_3^2+x\_4^2+x\_1x\_2+x\_2x\_3+x\_3x\_4+x\_4x\_1+x\_1x\_3+x\_2x\_4+x\_3x\_1+x\_4x\_2,
$$
again under the assumption that $x\_1+x\_2+x\_3+x\_4=4$. This is also minimized for the values $x\_1=x\_2=x\_3=x\_4=1$.


Is there a simple explanation for this? Is there a simple argument showing that the same will happen for, say, 12 variables and correlations of order up to 3?



Minimizing $x_1^2+x_2^2+x_3^2+x_1x_2+x_2x_3+x_3x_1$
The question is about the signature of a quadratic form
$$
\sum\_{i=1}^n x\_i^2 + \frac12\sum\_{1 \le \mathrm{dist}(i,j) \le p} x\_ix\_j
$$
(or about the spectrum of the corresponding linear operator). The matrix is a circulant matrix, see [Wikipedia](https://en.wikipedia.org/wiki/Circulant_matrix), and its eigenvectors are
$$
v\_k = (1, \omega^k, \ldots, \omega^{k(n-1)}),
$$
where $\omega = e^{\frac{2\pi i}{n}}$. The eigenvalues are easy to compute:
$$
\lambda\_k = \frac12 + \frac12 \sum\_{i=-p}^{i = p} \omega^{ki} = \frac12 + \frac12 \frac{\omega^{\frac{(2p+1)k}{2}} - \omega^{\frac{-(2p+1)k}{2}}}{\omega^{\frac{k}2} - \omega^{-\frac{k}2}} = \frac12 + \frac12 \frac{\sin \frac{(2p+1)k\pi}{n}}{\sin \frac{k\pi}{n}}.
$$
Unfortunately, there may be negative eigenvalues. For example, for $n=7$, $p=2$ we have
$$
\lambda\_5 = \frac12 + \frac12 \frac{\sin\frac{25\pi}7}{\sin\frac{5\pi}7} < 0.
$$


Thus for seven variables and cyclic sums up to order two the minimum will not be attained at $x\_1 = \cdots =x\_7 = 1$.


