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Let $w=v^{\prime }-v$ and $D(v)=\mathrm{sym}\nabla v$. The supporting-hyperplane inequality for the convex viscous dissipation potential givesThe stress is symmetric because its argument is a symmetric rate-of-strain tensor, so ${\sigma }_{ij}(D_{ij}^{\prime }-D_{ij})={\sigma }_{ij}{w}_{i,j}$. Subtracting the two functional values yieldsIntegration by parts turns the first integral intoInertialess 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 principleOnly 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 $(-pI):(D^{\prime }-D)=-p\mathrm{div}w=0$; the same proof applies to the deviatoric constitutive law.