Volume of balls in 3-dimensional manifolds with nonpositive Ricci tensor The volume of a ball in a 3-dimensional Riemannian manifold with nonpositive Ricci tensor is greater or equal to the volume of an Euclidean ball with the same radius?
It should not be true, but I am not finding a counterexample. In dimension larger than 3, a zero-Einstein nonflat manifold should be a counterexample, since by Bishop theorem it would follow that the manifold is flat.
 A: If you are OK with considering large balls, there are easy counterexamples. For example $T^2 \times \mathbb{R}$. Alternatively, there is a metric of negative Ricci curvature on $S^3$ (I think originally proven by Gao and Yau and Brooks, there is also later work by Lohkamp).
However, below the injectivity radius this seems to be true! I found this proof in a paper of Gimeno, but I sort of suspect that something like this was known in an earlier form.
The proof is cute so I record it here:
Let $S_r(p)$ denote the sphere of radius $r$ at a distance from $p$, for $r<\textrm{inj}(p)$. This forms a unit speed family of surfaces. Let $A(r)$ denote the area of $S_r(p)$. Note that $V(r) = \int_0^r A(t) dt$. Furthermore,
$$
A'(r) = \int_{S_r} H 
$$
and
$$
A''(r) = \int_{S_r} H^2 - \textrm{Ric}(\nabla r,\nabla r) - |h|^2
$$
where $h$ is the second fundamental form of $S_r$. The traced Gauss equations yield
$$
R = 2K + 2\textrm{Ric}(\nabla r,\nabla r) + |h|^2 - H^2. 
$$
We use this to eliminate the $|h|^2$ term.
\begin{align*}
A''(r) & = \int_{S_r} H^2 - \textrm{Ric}(\nabla r,\nabla r) - R + 2K + 2\textrm{Ric}(\nabla r,\nabla r) - H^2\\
& = \int_{S_r}  \textrm{Ric}(\nabla r,\nabla r) - R + 2K \\
& = \int_{S_r} - \textrm{tr}_{T\Sigma}\textrm{Ric} + 2K\\
& \geq 8\pi
\end{align*}
Integrating this from $0$ we find $A(r)  \geq 4\pi r^2$ so $V(r) \geq \frac{4\pi}{3} r^3$.
