= Solution
For a uniform classical <spin> $S\mathbf n$ with $\mathbf n^2=1$, the <energy> per bond is
$$
e=-S^2[J^x n_x^2+J^y n_y^2+J^z n_z^2].
$$
The strict inequalities among the couplings make the $x$ direction the unique easy axis. Thus
$$
\boxed{\mathbf S_j=+S\hat{\mathbf x}\ \hbox{for every }j,
\quad\hbox{or}\quad\mathbf S_j=-S\hat{\mathbf x}\ \hbox{for every }j,\qquad E_0=-LJ^xS^2.}
$$
These uniform configurations are also global classical minima: the largest singular value of the diagonal exchange matrix is $J^x$, so each bond has energy at least $-J^xS^2$, and both configurations saturate that bound.
A global <rotation> by $\pi$ about the $z$ axis sends $(S^x,S^y,S^z)$ to $(-S^x,-S^y,S^z)$, preserves every exchange product, and exchanges the two <ground states>. A $\pi$ rotation about $y$ does the same. The broken discrete symmetry can be represented by either operation.
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