= Solution
Put $r=|\mathbf x|$. In the <Unscaled Papkovich–Neuber representation>, choose the <componentwise harmonic vector field>
$$
\boldsymbol\Phi=-\tfrac12f(r)\boldsymbol\Omega\times\mathbf x,
\qquad f(r)=A+\frac B{r^3}.
$$
Both $\boldsymbol\Omega\times\mathbf x$ and $(\boldsymbol\Omega\times\mathbf x)/r^3$ are harmonic away from the origin: the latter is a linear combination of derivatives of the <harmonic function> $1/r$. Also $\mathbf x\cdot\boldsymbol\Phi=0$, so the representation gives $\mathbf u=f(r)\boldsymbol\Omega\times\mathbf x$. The <no-slip boundary condition> requires $f(a)=1$ and $f(b)=0$, giving
$$
A=-\frac{a^3}{b^3-a^3},\qquad B=\frac{a^3b^3}{b^3-a^3}.
$$
Thus the <Rotational Stokes flow between concentric spheres> is
$$
\boxed{\mathbf u=\frac{a^3}{b^3-a^3}\left(\frac{b^3}{r^3}-1\right)
\boldsymbol\Omega\times\mathbf x.}
$$
The <divergence> of $\boldsymbol\Phi$ is zero: the cross-product field itself has zero <divergence>, and $\nabla f$ is radial and perpendicular to $\boldsymbol\Omega\times\mathbf x$. Hence $p=2\mu\nabla\cdot\boldsymbol\Phi=0$ in this <pressure> gauge.
There is also a symmetry argument. Take the rotation axis as the polar axis. <Linearity of Stokes flow> makes <velocity> and <pressure> perturbations change sign when the rotation reverses. Reflection in any plane containing that axis reverses the axial rotation vector. Combined with axial symmetry, this excludes meridional <velocity> components and forces the <pressure> perturbation to be both unchanged and sign-reversed by that reflection. Equivalently a purely azimuthal axisymmetric <Stokes flow> has no meridional component of its vector <Laplacian>, so its meridional <pressure> gradient vanishes. Thus the <pressure> is constant; setting that arbitrary constant to zero gives the same result without solving the radial coefficients.
For the <Newtonian fluid stress tensor>, let $\mathbf q=\boldsymbol\Omega\times\mathbf x$. The symmetric part of the <velocity> gradient is
$$
e_{ij}=\frac{f'(r)}{2r}(q_ix_j+x_iq_j),
$$
because the gradient of rigid rotation is antisymmetric. Consequently the full <stress tensor>, with $p=0$, is
$$
\boxed{\sigma_{ij}=-\frac{3\mu B}{r^5}(q_ix_j+x_iq_j).}
$$
For $\boldsymbol\Omega=\Omega\mathbf e_z$, its only nonzero spherical-coordinate components are $\sigma_{r\phi}=\sigma_{\phi r}=-3\mu B\Omega\sin\vartheta/r^3$. On the inner sphere the traction exerted by the fluid, using the sphere's outward normal $\mathbf n$, is
$$
\mathbf t=\boldsymbol\sigma\mathbf n
=-\frac{3\mu B}{a^3}\boldsymbol\Omega\times\mathbf n.
$$
Its <torque> is
$$
\int_{r=a}\mathbf x\times\mathbf t\,dS
=-3\mu B\int\{\boldsymbol\Omega-\mathbf n(\mathbf n\cdot\boldsymbol\Omega)\}\,d\omega
=-8\pi\mu B\boldsymbol\Omega,
$$
using $\int n_in_j\,d\omega=4\pi\delta_{ij}/3$. Therefore the applied <Torque in rotational Stokes flow between concentric spheres> is
$$
\boxed{\mathbf G_0=\frac{8\pi\mu a^3b^3}{b^3-a^3}\boldsymbol\Omega.}
$$
For $b\gg a$, the relevant speed and length are $\Omega a$ and $a$, so negligible inertia requires the rotational <Reynolds number>
$$
\boxed{\frac{\rho|\Omega|a^2}{\mu}\ll1.}
$$
For a narrow gap $d=b-a\ll a$, the viscous gradients are controlled by $d$, not $a$. A centrifugal acceleration of scale $\Omega^2a$ is then compared with viscous scale $\mu\Omega a/d^2$, rather than $\mu\Omega/a$. Their ratio is $\rho|\Omega|d^2/\mu$. This explains why the large-gap condition is not a necessary narrow-gap criterion; a thin gap greatly strengthens viscous resistance. This base-flow scaling is distinct from a criterion for stability to arbitrary short-scale disturbances.
The <Minimum-dissipation theorem for Stokes flow> compares a Stokes solution with all sufficiently regular incompressible trial velocities in the same fluid domain that have the same prescribed boundary velocities. With no body-force work, the dissipation functional is
$$
D[\mathbf v]=\int 2\mu\,\mathbf e(\mathbf v):\mathbf e(\mathbf v)\,dV.
$$
If $\mathbf v=\mathbf u+\mathbf w$ and $\mathbf w=0$ on the prescribed boundary, integration by parts with $\nabla\cdot\boldsymbol\sigma(\mathbf u)=0$ and $\nabla\cdot\mathbf w=0$ makes the cross term vanish. Thus
$$
D[\mathbf v]-D[\mathbf u]=2\mu\int\mathbf e(\mathbf w):\mathbf e(\mathbf w)\,dV\ge0.
$$
The comparison fixes velocities, rather than applied torques or forces.
Now let the particles take their actual rigid translations and rotations in the particle-containing flow. Extend the fluid <velocity> through each particle by its rigid <velocity>. <No-slip boundary conditions> make this extension continuous; rigid motion has zero <rate-of-strain tensor> and zero <divergence>. The extension is therefore an admissible trial field in the original particle-free annulus, and its dissipation equals that of the fluid outside the particles. Hence $D\ge D_0$.
The Stokes boundary-work identity gives $D=\mathbf G\cdot\boldsymbol\Omega$. The outer sphere is stationary, the maintained inner sphere has zero translational <velocity>, and every added particle has zero total <force> and <torque>, so none of those other boundaries contributes power. Likewise $D_0=\mathbf G_0\cdot\boldsymbol\Omega$. Therefore
$$
\boldsymbol\Omega\cdot\mathbf G\ge\boldsymbol\Omega\cdot\mathbf G_0.
$$
For nonzero rotation and a particle of positive volume the inequality is strict: equality in the dissipation identity would <force> the extension to be the unique original flow, but that original flow has nonzero strain on every open particle region, whereas a rigid interior has zero strain. This proves that <force-free inclusions increase rotational resistance>:
$$
\boxed{\mathbf G\cdot\frac{\boldsymbol\Omega}{|\boldsymbol\Omega|}
>\frac{8\pi\mu a^3b^3}{b^3-a^3}|\boldsymbol\Omega|.}
$$
No dilute-suspension approximation or assumption about the particle shapes was used.
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