 Solution

A relative angle $qc$ between neighbors gives in-plane dot products $S^2\cos (qc)$ and $S^2\cos (2qc)$ at the two ranges. Set $x=\cos (qc)\in [-1,1]$. ThenThe classical helical minimum of the J1-J2 XY chain follows because for $J_2<0$ this is a strictly convex quadratic in $x$. Its unconstrained minimum isIt is a genuine noncollinear helix when $|J_2|>|J_1|/4$. Both chiralities $q$ and $-q$ have the same energy, and the pitch is defined modulo $2\pi /c$. At equality the helix merges into a collinear endpoint, so the strict inequality is required for an energy strictly lower than both collinear choices.

For a direct comparison write $K=|J_2|$. The minimum and the energy differences areBoth differences are strictly positive in the stated interior region. The stronger classical global argument Fourier-transforms the translationally invariant bilinear exchange: its energy is minimized by the wave vector maximizing $J_1\cos (qc)+J_2\cos (2qc)$, and the planar single-$q$ helix realizes the fixed spin length at every site. This establishes the minimum within the classical/mean-field model, not an assertion of true long-range magnetic order at positive temperature in a one-dimensional short-range quantum chain.