Suppose $f(x)$ is a convex and self-concordant function minimized at $x^*$. I have two starting points $\tilde{x}_0$ and $\hat{x}_0$ such that $|\hat{x}_0-x^*| \le |\tilde{x}_0 - x^*|$. We also know that $sign(\hat{x}_0 - x^*) = sign(\tilde{x}_0 - x^*)$. This obviously means that $f(\hat{x}_0) \le f(\tilde{x}_0)$.
We perform one Newton step to each initial point, obtaining $\hat{x}_1 = \hat{x}_0 - f'(\hat{x}_0)/f''(\hat{x}_0)$ and $\tilde{x}_1 = \tilde{x}_0 - f'(\tilde{x}_0)/f''(\tilde{x}_0)$. Will we still have $|\hat{x}_1-x^*| \le |\tilde{x}_1 - x^*|$ and thus $f(\hat{x}_1) \le f(\tilde{x}_1)$?