Assume Sobolev embedding： $$H_{1}^{1}(M) \subset L^{n /(n-1)}(M)$$ holds for the compact Riemannian manifold $(M^n,g)$, then we can get
$$\operatorname{Vol}_{g}\left(B_{x}(r)\right)^{(n-1) / n} \leq C \frac{d}{d r} \operatorname{Vol}_{g}\left(B_{x}(r)\right)+C \operatorname{Vol}_{g}\left(B_{x}(r)\right) \tag{2}$$ for every $x\in M$ and almost every $r>0$
Prove :
$$\frac{1}{2 C} \mathrm{Vol}_{g}\left(B_{x}(r)\right)^{1-1 / n} \leq \frac{d}{d r} \mathrm{Vol}_{g}\left(B_{x}(r)\right)\tag{3}$$
For $r<R$, If $\mathrm{Vol}_{g}\left(B_{x}(R)\right) \le (1/2C)^n$
Moreover prove the volumn of the ball has the lowerbound:
$$\operatorname{Vol}_{g}\left(B_{x}(R)\right) \geq \min \left(\left(\frac{1}{2 C}\right)^{n},\left(\frac{R}{2 n C}\right)^{n}\right)\tag{4}$$
For almost all $R>0$.
I have no idea how to deduce $(3)$ from $(2)$ .I want to use absorbing inequality but I am not sure if it's applicable?
Let $r<R$. Then $\mathrm{Vol}_{g}(B_{x}(r)) \leq \mathrm{Vol}_{g}(B_{x}(R)) \leq (1/2C)^{n}$. Hence $\mathrm{Vol}_{g}(B_{x}(r)) \leq (1/2C)^{n}$, which is equivalent to $C \leq \mathrm{Vol}_{g}(B_{x}(r))^{-1/n}/2$. Using (2) and this, we find \begin{align*} \operatorname{Vol}_{g}\left(B_{x}(r)\right)^{(n-1) / n} &\leq C \frac{d}{d r} \operatorname{Vol}_{g}\left(B_{x}(r)\right)+C \operatorname{Vol}_{g}\left(B_{x}(r)\right) \\ & \leq C \frac{d}{d r} \operatorname{Vol}_{g}\left(B_{x}(r)\right)+\frac{\mathrm{Vol}_{g}(B_{x}(r))^{-1/n}}{2} \operatorname{Vol}_{g}\left(B_{x}(r)\right) \\ & =C \frac{d}{d r} \operatorname{Vol}_{g}\left(B_{x}(r)\right)+\frac{1}{2} \operatorname{Vol}_{g}\left(B_{x}(r)\right)^{(n-1)/n}. \end{align*} This shows (3).