= Solution
Let $w=v'-v$ and $D(v)=\operatorname{sym}\nabla v$. The supporting-hyperplane inequality for the convex <viscous dissipation potential> gives
$$
\Omega(D')-\Omega(D)\geq\sigma_{ij}(D'_{ij}-D_{ij}).
$$
The <stress> is symmetric because its argument is a symmetric <rate-of-strain tensor>, so $\sigma_{ij}(D'_{ij}-D_{ij})=\sigma_{ij}w_{i,j}$. Subtracting the two functional values yields
$$
\mathcal F(v')-\mathcal F(v)
\geq\int_{\mathcal D}\sigma_{ij}w_{i,j}\,dx
-\int_{\mathcal D}\rho g_iw_i\,dx
-\int_{S_t}t_i^0w_i\,dS.
$$
<Integration by parts> turns the first integral into
$$
\int_{\partial\mathcal D}\sigma_{ij}n_jw_i\,dS
-\int_{\mathcal D}\sigma_{ij,j}w_i\,dx.
$$
Inertialess <force balance> is $\sigma_{ij,j}+\rho g_i=0$. The variation vanishes on $S_v$, and the <traction> is $t_i^0$ on $S_t$. Thus both the volume and boundary contributions cancel exactly, proving the <convex viscous potential minimum principle>
$$
\boxed{\mathcal F(v')\geq\mathcal F(v)
\quad\text{for every admissible }v'.}
$$
Only convexity, not strict convexity, is required, so no uniqueness follows just from this argument. For <incompressible flow>, admissible trial <velocities> must also be divergence free. The <pressure> term then does no work, because $(-pI):(D'-D)=-p\,\operatorname{div}w=0$; the same proof applies to the deviatoric constitutive law.
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