# Counterexample to Gunther Theorem when assuming only a Ricci curvature upper bound

One of Gunther's comparison theorems states that if $(M,g)$ is a complete riemannian manifold, $m\in M$ , $\mathbb{B}_r(m)$ is a (geodetic) ball which does not touch Cut$_m$ and there exists a $b\in\mathbb{R}$ such that $K\le b$ (where $K$ is the sectional curvature) we can say that $$vol_g(\mathbb{B}_r(m))\ge V^b(r)$$ where $V^b(r)$ is the volume of a ball of radius $r$ in the space form of curvature $b$.

I would like to find a counterexample as to why this statement cannot be true if we were to assume the same hypothesis but the one about sectional curvature, replacing it with one about Ricci curvature ($Ric_g \le (n-1)bg$ ). I am aware that it can be shown that every riemannian manifold (under suitable assumptions) admits a metric whose Ricci curvature is strictly negative, though I have not yet managed to see how it can be useful in this context. The problem which I encounter is that volume is a Riemannian notion, while I assume the existence of metrics of negative curvature is useful to gather differential or topological informations.

For instance $X:=S^2(1)\times S^2(1),\ Y:=S^4(\sqrt{3})$ They are Einstein. In space form of curvature $K$, $$g_K=dr^2+ \frac{1}{\sqrt{K}}\sin\ \sqrt{K}r\ g_{S^{n-1}}$$
$r$-geodesic ball in $X$ has an upper bound $4\pi^2(1-\cos\ r)^2$. However, in $Y$, $r$-geodesic ball has volume $6\pi^2 (1-\cos\ r/\sqrt{3})$.