By Nash embedding theorem, any complete Riemannian manifold $M$ can be isometrically embedded in $\mathbb{R}^N$, for sufficiently large $N$.

The proof of the theorem is quite involved, and it is not clear to me how the mean curvature of the produced embedding depends on $M$.

My question is the following. Is there a way to build an isometric embedding with mean curvature bounded from above in terms of appropriate intrinsic geometric data of the original manifold (e.g. curvature bounds, injectivity radius, etc)?

An example of the result that I would hope is the following statement: There exists a constant $c_n>0$, depending only on $n$, such that for any complete $n$-dimensional Riemannian manifold $M$ there exists $N>0$ and an isometric embedding of $M$ in $R^N$ such that the mean curvature of the embedding is not greater than $c_n \max\{K,1/\mathrm{i}\}$, where $K$ is the upper bound on the absolute value of the sectional curvatures of $M$ and $\mathrm{i}$ is a lower bound on the injectivity radius of $M$.

I am in particular interested in the case when $M = \mathbb{R} \times_f Z$ is the warped product of the euclidean $\mathbb{R}$ and a compact Riemannian manifold $(Z,h)$ with warping factor $f : I \to Z$.

I found plenty of lower bounds to the mean curvature of an isometric embedding in terms for so-called $\delta$-invariants defined by Chen (see e.g. here), but none of this results implies that an isometric immersion satisfying such a bound actually exists.

This question is somewhat related with this one.

EDIT: In the above, by "mean curvature" I mean the norm of the trace of the second fundamental form of the embedding. The second fondamental form is a symmetric tensor with values in the normal bundle.