As has already been pointed out, while the tubular radius bounds the curvature globally from below, the curvature information is not enough to correctly estimate this radius. Many examples can be considered, often silly ones when you allow your sub-manifold to be disconnected: consider the case of two parallel line segments in the plane and note that the embedding radius depends on their distance.

Since it is not clear what exactly you are after (computing the tubular radius of the curve $xy = 1$ is not so hard) or what information you already have, I would like to mention that your problem is solvable by calculus alone! Whether this calculus is tractable or not obviously depends on your choice of Riemannian manifold $X$ and submanifold $M \subset X$.

Given $M \subset X$, the *Medial Axis* $A_X(M)$ of $M$ in $X$ is defined to be the collection of all $x \in X \setminus M$ such that there are multiple solutions to the following constrained optimization problem in $X$:

**Minimize** $\text{dist}_X(x,m)$ **subject to** $m \in M$.

Here is a rough sketch of what a medial axis looks like when $X = \mathbb{R}^2$ and $M$ is the Nicolaescu horseshoe. The axis itself is in blue, and the red lines are my amateurish attempts at showing equidistant $M$-points

Once you know this medial axis, the distance from $A_X(M)$ to $M$ is precisely your tubular radius. Figuring out this distance again reduces to calculus which may be intractably hard depending on the choice of $X$ and $M$.

**Update** Here is a simple pictorial counterexample to the claim that if the curvature is decreasing from "left to right" then the tubular radius is the injectivity radius of the "left" endpoint. A straight line going up a lampshade suffices. If you want the curvature to decrease strictly, you can wrap the initial segment of the curve around the lower edge of the lampshade and then straighten as you go up. The point is that the curvature of the ambient manifold also plays a part in restricting the tubular radius. It is likely that your conjecture applies in Euclidean space, although I don't have an immediate proof of this.