= Solution
For unforced <incompressible flow>, multiply the <vorticity equation> by $\omega_i$. The antisymmetric part of the <velocity gradient tensor> drops out of $\omega_i\omega_j\partial_j u_i$, leaving
$$
\frac{D}{Dt}\frac{\omega^2}{2}=\omega_i\omega_jS_{ij}+\nu\omega_i\nabla^2\omega_i.
$$
<Statistical homogeneity> removes mean divergences. <Integration by parts>, together with $\nabla\cdot\boldsymbol\omega=0$, gives
$$
\boxed{\frac{d}{dt}\frac{\langle\omega^2\rangle}{2}
=\langle\omega_i\omega_jS_{ij}\rangle-\nu\langle|\nabla\boldsymbol\omega|^2\rangle
=\langle\omega_i\omega_jS_{ij}\rangle-\nu\langle|\nabla\times\boldsymbol\omega|^2\rangle}.
$$
The last equality holds for the averages, not pointwise. It follows from the usual <curl>/divergence identity after the divergence terms have averaged to zero. <Vortex stretching> produces <enstrophy>, while viscous diffusion destroys it.
Let $T=\ell/u$, $\epsilon\sim u^3/\ell$, and use the <Kolmogorov microscales> $\eta=(\nu^3/\epsilon)^{1/4}$ and $\tau_\eta=(\nu/\epsilon)^{1/2}$. Since $\langle\omega^2\rangle=\epsilon/\nu$, a slowly changing mean <enstrophy> has a time derivative of order $\epsilon/(\nu T)$. The usual small-scale estimate gives
$$
\nu\langle|\nabla\times\boldsymbol\omega|^2\rangle
\sim\frac{\epsilon}{\eta^2}
=\left(\frac{\epsilon}{\nu}\right)^{3/2}.
$$
The relative imbalance is consequently $\tau_\eta/T\sim(\nu/u\ell)^{1/2}=\mathrm{Re}^{-1/2}$, yielding
$$
\boxed{\langle\omega_i\omega_jS_{ij}\rangle
=\nu\langle|\nabla\times\boldsymbol\omega|^2\rangle[1+O(\mathrm{Re}^{-1/2})]}.
$$
The estimate assumes an established high-Reynolds-number cascade and integral-time evolution of the mean, not a rapid initial transient. In a forced flow the exact balance includes $\langle\boldsymbol\omega\cdot\nabla\times\mathbf f\rangle$; forcing smooth on the <integral scale of turbulence> is smaller than the leading small-scale terms.
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