Converse to Bishop-Gromov Inequality Is the converse to the Bishop-Gromov Inequality true? 
In other words, if, for a complete $n$-dimensional Riemannian manifold $M$, there is $k \in \mathbb{R}$, such that defining $V_k(R)$ to be the size of a ball of radius $R$ in the standard space $S^n_k$, for any $p \in M$ and $R\geq 0$ we have that 
$$
vol_M(B_R(p)) \leq V_k(R)
$$
then is it true that on $M$, $Ric \geq (n-1)k$?
 A: This is false for $n=3$, and for $k<0$. One can get a different pinching condition, $R(g)g-Ric \geq -4g$ which is weaker than $Ric \geq -2g$ ($R(g)$ is the scalar curvature), and which gives the same upper bound on volume as Gromov's inequality (for a manifold with negative curvature or no cut points). The idea is that in the proof of Gromov's inequality, one can integrate out the curvature tangent to the sphere of radius $r$ using the theorem egregrium and Gauss-Bonnet, giving a weaker curvature condition. There are metrics which satisfy the first curvature condition but not the Ricci lower bound. For example, there are homogeneous negatively curved metrics on $R^3$ which in one direction are foliated by totally geodesic hyperbolic planes of curvature $-K_1^2$, and in a perpendicular direction are foliated by totally geodesic hyperbolic planes of curvature $-K_2^2$ for $0 < K_1\neq K_2>0$ (this is a double warped product, and is Thurston's "ninth geometry" of the form $dr^2+e^{2K_1r}dx^2+e^{2K_2r}dy^2$). One obtains a counterexample for the choice of $K_1 =5/4,K_2=1/2$.   
Added computations: The curvature operator of this metric may be normalized to have sectional curvatures $-K_1^2, -K_2^2, -K_1K_2$. If we assume that $K_1\geq K_2$, then the minimal curvature is $-K_1^2$. Under the same assumption, the minimal eigenvalue of the Ricci operator is $-K_1^2-K_1K_2$. The minimal eigenvalue of $Rg-Ric$ is $-2K_1^2-K_1K_2-K_2^2$. We then see that for $K_1=5/4, K_2=1/2$, the minimal eigenvalue of $Ric$ is $-35/16<-2$, and the minimal eigenvalue of $Rg-Ric$ is $-4$. 
A: If it is true for all $R,$ then the inequality follows from the infinitesimal calculation (as $R$ goes to $0.$)
