min max of a function and integral of the its gradient Suppose $\mathcal{M}$ is a Riemannian manifold with metric $g$ and with the volume measure $d\mu$ (induced by $g$). Let $f:\mathcal{M}\to \mathbb{R}$ be $C^{\infty}$. Is it true that
$$\max_{\mathcal{M}}f-\min_{\mathcal{M}}f\leq\int_{\mathcal{M}}|\nabla f|d\mu \;?$$
Here $|\nabla f|$ is the norm of the gradient of $f$ with respect to the metric $g.$
Edit 1:
Following the comments I'd like to know if the following inequalities hold 
$$(\max_{\mathcal{M}}f-\min_{\mathcal{M}}f)\operatorname{diam}{({\mathcal{M}})}\leq\int_{\mathcal{M}}|\nabla f|d\mu \;?$$
Edit 2:
$$(\max_{\mathcal{M}}f-\min_{\mathcal{M}}f)\operatorname{Vol}{(\mathcal{M})}\leq\int_{\mathcal{M}}|\nabla f|d\mu \;?$$
 A: No. 
Let $M = (0,1)\times (0,\epsilon)\subset \mathbb{R}^2$ with coordinates $(x,y)$ and with the metric inherited from the Euclidean metric. Take $f(x,y) = x$. $|\nabla x| = 1$. $\sup f - \inf f = 1$ (Please note that in general on an arbitrary set $S$ given a function $f:S\to\mathbb{R}$, the notion $\max_S f$ is generally not defined, since you cannot guarantee that $f$ will attain its maximum on $S$). But $\int_M |\nabla f| d\mu = \epsilon$ which can be made arbitrarily small. 

Response to edits 1 & 2


*

*My counter example above is still a counterexample for Edit 1. 

*For a counter example to edit 2: observe that it is false even in one dimension. Let $f$ be an arbitrary non-negative bump function supported on (-1,1). Consider the manifolds $M_n = (-n,n)$. The RHS stays constant, but the LHS can grow to as large as you want. 


Remark: What you are writing down is getting closer and closer to Sobolev/Poincare inequalities.  For more details on the scaling/dimensional analysis argument, you may want to read Terry's answer here and his linked blog post. 
