= Solution
Double contraction of the projected <Gauss equation>, with the timelike-normal sign and curvature convention in the question, gives
$$
R^{(3)}=q^{\mu\rho}q^{\nu\sigma}R^{(4)}_{\mu\nu\rho\sigma}-K^2+K_{\mu\nu}K^{\mu\nu}.
$$
Here $q^{\mu\nu}=g^{\mu\nu}+n^\mu n^\nu$. Expanding the two projectors gives
$$
q^{\mu\rho}q^{\nu\sigma}R^{(4)}_{\mu\nu\rho\sigma}
=R^{(4)}+2R^{(4)}_{\mu\nu}n^\mu n^\nu.
$$
The four-normal contraction vanishes by Riemann antisymmetry. Because $g_{\mu\nu}n^\mu n^\nu=-1$, the right side is $2G^{(4)}_{\mu\nu}n^\mu n^\nu$. Rearranging proves the <Hamiltonian constraint> identity:
$$
\boxed{G^{(4)}_{\mu\nu}n^\mu n^\nu=\frac12\left[R^{(3)}+K^2-K_{\mu\nu}K^{\mu\nu}\right].}
$$
For flat FRW, $R^{(3)}=0$, $K^2=9H^2$ and $K_{\mu\nu}K^{\mu\nu}=3H^2$, so $G_{nn}=3H^2$. Einstein's equation is therefore the first <Friedmann equation>, $3H^2=8\pi G\rho$, or $3H^2=8\pi G\rho+\Lambda$ if a cosmological constant is placed separately in Einstein's equation.
The scalar $R^{(3)}$ measures intrinsic spatial curvature, $K^2$ measures squared volume expansion, and the norm of $K_{ij}$ measures directional expansion including shear. More generally $K_{ij}=Kq_{ij}/3+s_{ij}$ with trace-free $s_{ij}$ gives $K_{ij}K^{ij}=K^2/3+s_{ij}s^{ij}$. The shear vanishes in FRW, leaving the positive $3H^2$ contribution. This is an instantaneous constraint on initial data, not the separate acceleration equation.
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