= Solution
When both <spheres> have the same effective downward weight, the second <sphere> adds a force-driven disturbance instead of being merely a force-free rigid inclusion. The power now contains both $WV_{1z}$ and $WV_{2z}$, so the one-forced-sphere bound from part (iv) no longer applies.
For $a\ll R$, place the <spheres> in a tandem configuration with $a\ll s\ll R$. Locally the flow around the second <sphere> is dominated by its downward <Stokeslet>. Its axial incident <velocity> at the first <sphere> is downward, so the first <sphere> falls faster to leading order. By contrast, the <zero total flux for localized tube Stokes flow> requires compensating upward flow away from the downward-moving particle core in sections cutting the <sphere>. There are therefore positions in the return flow where the disturbance acts upwards.
One can choose the second <sphere> in such an upward-flow region generated by the first. The <Lorentz reciprocal theorem> makes the axial cross coefficients of the <hydrodynamic mobility matrix> equal, so the downward <force> on the second <sphere> then produces an upward contribution to the first <sphere>'s <velocity>. With small well-separated <spheres>, this force-driven interaction dominates the higher-order self-resistance change from simply inserting the other <sphere>. In mobility notation, with downward positive,
$$
V_{1z}\simeq V_0+W\mathcal M_{12}^{zz},
$$
and the cross coefficient can have either sign in the tube. Thus
$$
\boxed{V_{1z}>V_0\text{ in a downward entraining configuration},\qquad
V_{1z}<V_0\text{ in a return-flow configuration}.}
$$
The distinction is the sign of the incident flow, not whether adding a second geometrical obstacle lowers a fixed-velocity dissipation.
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