= Solution
Use the unitary discrete <Fourier transform> $a_j=L^{-1/2}\sum_ke^{ikj}a_k$, where $k=2\pi n/L$. Then
$$
H_{\rm LSW}=\sum_k\left[\alpha_k a_k^\dagger a_k+
\frac{\beta_k}{2}(a_ka_{-k}+a_k^\dagger a_{-k}^\dagger)\right],
$$
with $\alpha_k=S[2J^x-(J^y+J^z)\cos k]$ and $\beta_k=-S(J^y-J^z)\cos k$. The factor $1/2$ accounts for the duplicated $k,-k$ pair in the full momentum sum.
Make a bosonic <Bogoliubov transformation> $a_k=\cosh\theta_k\,b_k+\sinh\theta_k\,b_{-k}^\dagger$. Its coefficients preserve the <canonical commutation relation> because $\cosh^2\theta_k-\sinh^2\theta_k=1$. The coefficient of the pair term vanishes when $\tanh2\theta_k=-\beta_k/\alpha_k$. Thus
$$
\boxed{H_{\rm LSW}=\sum_k\varepsilon_k b_k^\dagger b_k+\text{constant},\qquad
\varepsilon_k=2S\sqrt{(J^x-J^y\cos k)(J^x-J^z\cos k)}.}
$$
The dropped constant is $\tfrac12\sum_k(\varepsilon_k-\alpha_k)$ in addition to the classical <ground-state energy>. Since $J^x>J^y>J^z>0$, $\alpha_k>|\beta_k|$ at every $k$, so the <bosonic Bogoliubov diagonalization> is stable. The minimum occurs at $k=0$ and gives the <energy gap> $2S\sqrt{(J^x-J^y)(J^x-J^z)}$. In the isotropic limit the pair coefficient vanishes and the familiar quadratic small-$k$ ferromagnetic <dispersion relation> is recovered.
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