Suppose that all eigenvalues of the second fundamental form $A=\{h_{ij}\}$ of manifold $M$ are strictly positive, then there is some $\epsilon>0$ such that the inequality $$h_{ij}>\epsilon Hg_{ij}$$ holds everywhere on M, where $H=tr_{g}A$
My proof is as follows:
select $\epsilon$ such that $1>n\epsilon$. Since $g^{ij}g_{ij}=n$, we have $1>g^{ij}g_{ij}\epsilon$. Since all eigenvalues are strictly positive, $H$ is strictly positive as well as the trace of the matrix is equal to the sum of eigenvalues. Thus, we can multiply $1>g^{ij}g_{ij}\epsilon$ by $H$: $$H>Hg^{ij}g_{ij}\epsilon.$$ Since $H=g^{ij}h_{ij}$, we have $$g^{ij}h_{ij}>Hg^{ij}g_{ij}\epsilon.$$ Then $$g^{ij}(h_{ij}-Hg_{ij}\epsilon)>0.$$ We thus conclude that $h_{ij}>\epsilon Hg_{ij}$.
The proof seems like a valid one. I cannot poke any holes in it. But it seems a bit ad hoc. I first took the inequality that I'm supposed to prove $h_{ij}>\epsilon Hg_{ij}$. Then took a trace of both sides, getting $H>\epsilon H n$, from which we get inequality $1> n\epsilon$. And then I proceeded in a somewhat backwards manner. I chose $\epsilon$ such that $1> n\epsilon$. The rest is above.
So I'm wondering if there is a proof such that inequality would follow from general differential geometric considerations rather than being a consequence of my rather ad hoc proof?
EDIT: I was wrong by concluding that $g^{ij}(h_{ij}-Hg_{ij}\epsilon)>0$ implies $h_{ij}>\epsilon Hg_{ij}$. But then I don't know how to finish my proof.
Fix a point $p\in M$. The second fundamental form at $p$ is a symmetric bilinear form, so by the spectral theorem there is an orthonormal basis $e_i$ for $T_p M$ made up of its eigenvectors; i.e. such that $A(e_i, e_j) = \lambda_i \delta_{ij} $. In this basis the matrix for $A$ is simply its eigenvalues on the diagonal and $g$ is the identity matrix, so we have $A \ge (\min_i \lambda_i) g$. From here (and the positivity assumption) we can easily get the pointwise inequality with $$\epsilon(p) = \frac12\frac{\min_i \lambda_i}{\sum_i \lambda_i}.$$
The remaining issue is whether this ratio is bounded away from zero so that we can choose $\epsilon = \inf_{p\in M} \epsilon(p)>0$. For non-compact $M$ it is possible that as we head off to infinity this ratio can approach zero - while I haven't worked out the details I think the surface of revolution of $e^{-x}$ is a counterexample. Thus you need some extra assumption - for example that the eigenvalues are bounded below by a positive constant, or compactness of $M$.