Let us say that two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a real vector space $V$ are strongly equivalent if there exists a constant $\lambda \geq 1$ such that $$ \frac{1}{\lambda} \left( \|x\|_1 + \|y\|_1 - \|x + y\|_1 \right) \leq \|x\|_2 + \|y\|_2 - \|x + y\|_2 \leq \lambda \left( \|x\|_1 + \|y\|_1 - \|x + y\|_1 \right) $$ for all vectors $x$ and $y$ in $V$.

A remark I owe to Suvrit is that if we take $y = -x/2$, this condition implies that $$ \frac{1}{\lambda} \|x\|_1 \leq \|x\|_2 \leq \lambda \|x\|_1 $$ so that the norms are equivalent in the usual sense.

Geometrically, two norms are strongly equivalent if the defect in the triangular inequality for any one of the norms is controlled by the defect of the other. In particular, both normed spaces must have exactly the same geodesics. Thus, for example, the $\ell_\infty$ and the $\ell_2$ norms on the plane are not strongly equivalent.

Question. Is there a simple criterion to determine whether two norms on the plane (or in a finite-dimensional space) are strongly equivalent. What if we assume that the unit spheres of both norms are polygons or that the unit spheres are smooth and positively curved?

Added 07/04/2014. Thanks in good measure to the exchange with Willie Wong, the answer to the OP in the two-dimensional case is the following simple (albeit surprising) result:

Proposition. Two norms $N_1$ and $N_2$ on the plane are strongly equivalent if and only if their distributional Laplacians, considered as measures on the unit circle, are comparable in the sense that there exists a constant $\lambda \geq 1$ such that $\lambda \Delta N_1 - \Delta N_2$ and $\Delta N_2 - \lambda^{-1} \Delta N_1$ are (positive) measures.

This contains basically all the geometric information one may wish (see my answer below).