If two Gaussians disagree on one moment, it seems like this should imply that they have a large variation distance--equivalently, if two Gaussians are close in variation distance it's hard for their moments to differ very much. But I'm having trouble proving this without getting into a terrible mess and ending up with weak bounds. I'm hoping someone here can offer a slick proof or point me to one.

Suppose that $\Sigma_1$ and $\Sigma_2$ are two positive-semidefinite matrices, with diagonal entries bounded by 1, and that $\Sigma_1$ and $\Sigma_2$ differ by at least $\delta$ in one of their entries.

Can we lower-bound the total variation distance between the multivariate gaussians with covariance matrices $\Sigma_1$ and $\Sigma_2$? In particular, is this distance at least $\Theta(\delta)$? It seems like it ought to be, since the gaussians are pretty well concentrated. But trying to formalize the obvious approach seems to create a bit of a mess.

Edit: Douglas Zare and guest point out that you can trivially reduce to the 2 dimensional case by projecting onto the relevant subspace. So we can restrict attention to that case.