 Solution

Let $s=⟨S_i^z⟩$ be uniform, with dimensionless spin operators $S=1/2$. The factor $1/2$ in the source Hamiltonian removes double counting of pairs. Decoupling ${\mathbf{S}}_i\cdot {\mathbf{S}}_j$ about a uniform polarization gives the one-site mean-field theory HamiltonianA body-centred-cubic site has eight nearest neighbors and six next-nearest neighbors, soFor a spin one half in this effective field, the mean-field self-consistency equation isLinearizing at the continuous onset gives $s=J_{0}s/(4k_{B}T)+O(s^3)$, henceEquivalently, the general spin-$S$ result is $k_B{T}_C^{\mathrm{MF}}=S(S+1)J_0/3$. The antiferromagnetic next-nearest-neighbor exchange reduces the uniform ordering scale rather than contributing its absolute magnitude. The negative nuclear g-factor does not change this exchange instability; aligning the spins also aligns their moments with one another.

For these particular competing couplings, the uniform mode is indeed the largest exchange eigenvalue. Writing $x=\cos (q_{x}a/2)$, $y=\cos (q_{y}a/2)$ and $z=\cos (q_{z}a/2)$ giveswith equality at $\mathbf{q}=0$ modulo reciprocal lattice vectors. To see the bound, replace a negative $xyz$ by its absolute value for an upper bound, let $t=(|xyz|{)}^{1/3}$, and use $x^2+y^2+z^2\ge 3t^2$. The expression is at most $8t^3-6t^2+3$, whose maximum on $0\le t\le 1$ is five. Thus another modulation does not preempt the stated ferromagnetic onset at this level of mean-field theory. Fluctuations can change the actual transition temperature.