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Keep the unit-normal sign explicit: $n^a{n}_a=\epsilon \in \{+1,-1\}$. Since the hypersurface is spacelike its normal is timelike. ConstructChoose the sign of n for the desired time orientation. The pullback of h is the nondegenerate induced metric. The second fundamental form isIts sign reverses when n reverses. It is tangential in both indices and symmetric: the antisymmetric part of the derivative of a normalized gradient vanishes after both tangential projections. Equivalently it is the projected Hessian of f divided by the normalizing factor. This is the normal-change definition of extrinsic curvature.

For tangent vector fields X,Y, let D be the induced Levi-Civita connection. Orthogonality to n gives the derivative decompositionUse the curvature convention of the preceding solution, $R(X,Y)Z=[{\nabla }_X,{\nabla }_Y]Z-{\nabla }_{[X,Y]}Z$. Substitute the decomposition twice and take its tangential component. Terms differentiated normally cancel, while the two normal components contribute the quadratic shape terms:Lowering the first curvature index gives the Gauss equation for a nonnull hypersurfaceAll four displayed indices on the left are tangential.

Contract with $h^{ac}{h}^{bd}$ and use $h^{ab}=g^{ab}-\epsilon n^a{n}^b$ for tangential contractions. The double-normal term vanishes by curvature antisymmetry and each mixed contraction is ambient Ricci curvature. With $K=h^{ab}{K}_{ab}$ this givesThis also records which metric and curvature conventions enter the sign.

In vacuum without a cosmological constant, the ambient Ricci tensor and scalar vanish. A totally umbilic hypersurface has $K_{ab}=\lambda h_{ab}$, so $K=3\lambda$ and $K_{ab}{K}^{ab}=3{\lambda }^2$. Therefore the scalar curvature of an umbilic vacuum hypersurface is $6\epsilon {\lambda }^2$.

To express the requested nonnegative version, take Lorentz signature $(+---)$, so the timelike unit normal has $\epsilon =+1$ and h is the induced negative-definite spatial metric. ThenIf instead one uses signature $(-+++)$ and its positive-definite spatial metric with the same curvature-slot convention, the scalar is $-6{\lambda }^2$. Replacing h by $-h$ reverses its scalar curvature. Thus the printed inequality has a signature/induced-metric convention built into it; the general formula above makes that choice explicit rather than losing the sign. The metric in the separate final question explicitly uses the mostly-plus signature.