This is not really an answer, but it's too long for a comment, so I'm posting it this way. One way in which the expression you are considering has appeared in recent years is via the heat equation shrinking plane curves. (See, for example, Gage and Hamilton, The heat equation shrinking convex plane curves, J. Differential Geometry 23 (1986), 69–96. They show that, in this case, one has
$$
\kappa_t = \kappa'' + \kappa^3 = \kappa^3\bigl(1 + \kappa''/\kappa^3\bigr).
$$
Whether this might be helpful to you, I don't know. It does appear to say that your inequality says something about how fast the curve will move under the heat shrinking flow.

As for a relation with centro-affine curvature, yes, this can be worked out. For abstract reasons, there is a polynomial relation among the quantities $\kappa$, $\kappa'$, $\kappa''$, $\lambda$, $\dot \lambda$, and $\ddot\lambda$, where $\lambda$ is the centro-affine curvature of the curve and the overset dot represents the derivative with respect to the centro-affine arclength $d\sigma$. I don't know a reference for this; it's classical (but messy). It starts with the easily-derived (and well-known) relation $\kappa (ds)^3 = \lambda (d\sigma)^3$ between the differentials and goes on from there. Actually, I think that the relation is better expressed between the quantities $\mu = \kappa^{1/3}$ and $\nu = \lambda^{1/3}$, but even this formula, when you work it out, isn't so nice.

Afterthought: It occurred to me that you put the 'centro' in parentheses because you might be interested in the affine curvature and the affine arclength, too. Of course, the affine arclength is $\kappa^{1/3}ds$, and the affine curvature $\tau$ is given by the formula
$$
\tau = \frac{\kappa\kappa'' + 3\kappa^4 - \tfrac{5}{3} (\kappa')^2}{3\kappa^{8/3}} .
$$
Thus, it's not quite what you want. As for a reference, you could consult any good differential geometry book that discusses affine geometry.