= Solution
Interpret the two displayed exchange components as belonging to the common nearest-neighbour sum with coefficient $-J$. Write $H_f=|h|$ and, for $h\ne0$, $s=\operatorname{sgn}h$. The source explicitly fixes $J>0$ but does not separately constrain $h$ in this question; keeping $H_f$ covers either <Magnetic field>[field] direction.
For <Vector>[classical vectors] of length $S$, define $\mathbf q_j=(S_j^x,S_j^y)$. The <Cauchy-Schwarz inequality> gives $\mathbf q_j\cdot\mathbf q_{j+1}\le|\mathbf q_j||\mathbf q_{j+1}|\le S^2$, and $hS_j^x\le H_fS$. Each term is therefore bounded below, so
$$
E_{\mathrm{cl}}\ge-LJS^2-LH_fS^2.
$$
Uniform alignment along the <Magnetic field>[field] saturates every bound:
$$
\boxed{\mathbf S_j=(sS,0,0),\qquad E_{\mathrm{cl,min}}=-L(J+H_f)S^2\quad(h\ne0).}
$$
Equality requires all planar <vectors> to be parallel with length $S$, and the nonzero <Magnetic field>[field] then fixes their direction. At $h=0$, every uniformly oriented <vector> in the $xy$ plane is a <Ground state>[classical ground state]:
$$
\boxed{\mathbf S_j=S(\cos\phi,\sin\phi,0),\qquad E_{\mathrm{cl,min}}=-LJS^2.}
$$
A nonuniform configuration cannot improve these simultaneously saturated bond bounds, including on the periodic ring.
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