I have a question about the last step in the proof of corollary 2.7 in the chapter 2 of Colding and Minicozzi's book A Course in Minimal Surfaces.
Corollary 2.7 If $\Sigma^2\subset\mathbb{R}^3$ is immersed and minimal, $B_{r_0}^{\Sigma}\subset \Sigma^2$ is an intrinsic disk of radius $r_0$, and $B_{r_0}^{\Sigma}\cap\partial\Sigma=\emptyset$, then $$something=2(Area(B_{r_0}^{\Sigma})-\pi r_0^2)\le Length(\partial B_{r_0}^{\Sigma})-2\pi r_0^2.$$
The proof of this last inequality is first to notice that $$\frac{d^2}{dt^2}Length(\partial B_t^{\Sigma})\ge 0$$ by a previous lemma, therefore $$t\frac{d}{dt}Length(\partial B_t^{\Sigma})\ge Length(\partial B_t^{\Sigma})$$
therefore
$$\frac{d}{dt}(Length(\partial B_t^{\Sigma})/t)\ge 0.$$
The authors say that the inequality that we want now follows easily. However, I cannot see why. Intuitively the length of $\partial B_t^{\Sigma}$ should be $c(t)\pi t$ and we now know $c(t)$ is increasing.
I guess then we should use the coarea formula somehow to finish it, but I just cannot write it out explicitly. Any help will be appreciated. Thanks in advance!
I integrated the equation:
$\int_0^{r_0} t\frac{d}{dt}Length(\partial B_t^\Sigma)\geq \int_0^{r_0}Length(\partial B_t^\Sigma)$
Observed that right hand side is equal to $Area(B_{r_0}^\Sigma)$
By integrating by parts left hand side we get:
$r_0Length(\partial B_t^\Sigma)- Area(B_{r_0}^\Sigma)\geq Area(B_{r_0}^\Sigma)$
and hence we conclude by subtracting from both equations $2\pi r_0^2$