Area of distance sphere in manifold with Ricci $\ge 0$. Let $M$ be a open complete manifold with Ricci curvature $\ge 0$.
By a theorem of Calabi and Yau, the volume growth of $M$ is at least of linear.
I am wondering whether the following statement is true:
Let $p$ be any fixed point in $M$ and $B(p, r)$ be the distance ball of radius $r$ in $M$. Then for any given $R>0$, there exists a constant $c=c(p,R)>0$ such that $Area(\partial B(p, r))\ge c(p,R)$ for any $r>R$.
 A: The answer is YES.
Let $b\colon M\to \mathbb R$ be a Busemann function for a ray $\gamma$ from $p$, so that $b(p)=0$ and $b(x)\le 0$ for any $x\in \gamma$.
Set 
$$L_t=b^{-1}(t)\ \ \text{and}\ \  L^-_t=b^{-1}(-\infty,t].$$ 
Note that the sublevel sets $L_t^-$ are mean curvature concave for all $t$.
In particular, any area minimizing hypersurface in $L_t^-$ with the boundary in $L_t$ lies in $L_t$.
Fix $t<0$.
From above, 
$$\mathop{\rm area}\partial B(p,r)\ge\mathop{\rm area}(\partial B(p,r)\cap L_t^-)\ge \mathop{\rm area}( B(p,r)\cap L_t)\ge \mathop{\rm area}( B(p,R)\cap L_t);$$
i.e., the inequality holds for $c(p,R)=\mathop{\rm area}( B(p,R)\cap L_t)$.
It remains to choose $t$ and $R$ so that $\mathop{\rm area}( B(p,R)\cap L_t)>0$; $R=2$ and $t=-1$ will do the job.
A: This is clearly false, just consider the cylinder 
$$ R_t \times  S_{\theta} $$
with the product metric 
$$g_\alpha=dt^2+\alpha^2 d\theta^2.$$  
This is a flat metric so $Ric_{g_\alpha} = 0$.  On the other hand, for $r>>\alpha$, it is easy to see $Area(\partial B_r)<8\pi \alpha$.  Since $\alpha$ is arbitrary there is no uniform lower bound.  
Maybe you need a uniform lower bound on the injectivity radius?
 (I'm not an expert on comparison geometry so don't know off the top of my head if this would suffice) [Edit: Or maybe this can only happen if the metric splits off an isometric euclidean factor].
[As an aside I can't seem to get math blackboard fonts to work anyone else have a problem with this?]
