= 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 $x=\cos(qc)\in[-1,1]$. Then
$$
\boxed{E_{\rm H}=-S^2[J_1\cos(qc)+J_2\cos(2qc)]
=S^2[J_2-J_1x-2J_2x^2].}
$$
The <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 is
$$
\boxed{x_*=-\frac{J_1}{4J_2},\qquad\alpha=\frac14.}
$$
It 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 are
$$
E_{\rm H,min}=-S^2\left(K+\frac{J_1^2}{8K}\right),\qquad
E_{\rm F}-E_{\rm H,min}=\frac{S^2(4K-J_1)^2}{8K},\qquad
E_{\rm AF}-E_{\rm H,min}=\frac{S^2(4K+J_1)^2}{8K}.
$$
Both 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.
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