I have been staring at this for far too long and am making no progress. Would someone with more geometry experiance mind taking a look?
Suppose $M \subset \mathbb R^d$ is a $C^2$ $(d-1)$-manifold. In particular I am interested in when $M$ is the set $\{x \in \mathbb R^d: f(x) \le 1\}$ for some $C^2$ function $f$. Suppose also the curvature at each point at least $m \ge 0$. There are loads of equivalent definitions of this. For example
For every twice-differentiable path $\gamma:(0,1) \to M$ with all $\|\gamma'(t)\| =1 $ we have $\|\gamma''(t)\| \ge m$ for all and $t \in (0,1)$. Here $\|\cdot\| $ is the Euclidean norm.
For each point $x \in M$ if we locally represent the manifold as the graph of a function $F:\mathbb R^{d-1} \to \mathbb R$ that is minimised at $x$ then the Hessian of $F$ at $x$ has all eigenvalues at least $m$.
Since the curvature describes the rate of change of the unit normal, we should be able to integrate something to see two points that are far away have normals that are far away.
Suppose $x,y \in M$ have unit normals $N_x$ and $N_y$. Is there a way to prove for example $\|N_x-N_y\| \ge Cm \|x-y\|$ for some constant $C\ge 0$ independent of $x,y$?
Trying to integrate something does not get me very far. Even assuming $\|\nabla f(x)\|=1$ and so the normal at $x$ is $\nabla f(x)$ I cannot do it.
For any $x,y$ let $\gamma:[0,1]$ be a geodesic from $y$ to $x$ with constant speed $s$ equal to the geodecis distance between the points. Since $\nabla f(x)$ has derivative $\nabla ^2f(x)$ some version of the fundamental theorem of algebra gives
$$\nabla f(x) - \nabla f(x) = \int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt $$
$$\implies \|\nabla f(x) - \nabla f(x)\|^2 = \left(\int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt \right) \cdot \left( \int_{0}^1 \nabla^2 f(\gamma(t)) \gamma'(t) dt \right) $$
what I would like to do now is say the right-hand side is at least $$ \int_{0}^1 \Big(\nabla^2 f(\gamma(t)) \gamma'(t) \Big) \cdot \Big (\nabla^2 f(\gamma(t)) \gamma'(t) \Big)dt $$ $$ = \int_{0}^1 \gamma'(t) \nabla^2 f(\gamma(t)) \nabla^2 f(\gamma(t)) \gamma'(t) dt.$$
By the Hessian condition the integrand should be at least $m^2\|\gamma'(t)\|^2 = m^2 s^2 \ge m^2 \|x-y\|^2$. Now take square roots to get $\|\nabla f(x)-\nabla f(y)\| \ge m \|x-y\|$.
Obviously there are loads of holes in the above. Does anyone know how to fix it or even if the fact I want to prove is true?
As you've probably noted, the main difficulty in obtaining this kind of estimate is that the inequalities you want to use (e.g. Cauchy-Schwarz) go the wrong way - it's easy to establish $N_x - N_y \lesssim x - y$ using an upper bound on curvature, but not the converse. Just establishing that $N$ moves quickly everywhere doesn't necessarily establish that it goes far, because (a priori) it could "loop around". Any argument needs to incorporate some global knowledge somehow. Here's how I'd prove your claim (or something very close):
Think of the unit normal as a map between manifolds $N : M \to S^{d-1},$ which we call the Gauss map, and recall that the second fundamental form is just $A = DN.$
Since you're assuming the curvature is strictly positive (your desired inequality is trivial in the case $m=0$), we know that $M$ is the boundary of a strictly convex set; so its normal vectors are unique and thus $N$ is a diffeomorphism. Thus we can study $\phi = N^{-1},$ for which our assumption $A \ge m$ implies $$D\phi = A^{-1} \le 1/m.$$ We can now apply an integration argument: given a geodesic $\gamma$ in $S^{n-1}$ joining $N_x$ to $N_y,$ we have
\begin{align} d(x,y) &= d(\phi(N_x), \phi(N_y)) \\ &\le \mathrm{len}(\phi \circ \gamma) \\ &= \int |D\phi ( \dot \gamma)| \\ &\le \int |D \phi| |\dot \gamma| \\ &\le \frac1m \int |\dot \gamma| \\ &= \frac1m d(N_x, N_y), \end{align}
i.e. $d(N_x, N_y) \ge m \, d(x,y).$