I can't understand the prove in [Xi-Ping Zhu] Lectures on mean curvature flows. The statement as follow.
Lemma 3.5 (page 32)
There exists a positive constant $C$ such that $$\textrm{sup}\{\overline{H}(x,\tau)|x\in M^n, r\geq 0 \}\leq C.$$
I can follow all the step but the last one. I don't know why
$$0\leq \frac{1}{S- \rho_0 }[2H^2-\rho_0|A|^2H]$$
at $(z_1,t_1)$ implies
$$H(z,t)\leq Cr_{out}(t_0)^{-1} ,$$ for all $z\in S^n, t\in[0,t_0].$
Here, $\Psi=\frac{H}{S-\rho_0}$ and $\Psi$ has its maximum at $(z_1,t_1)$
I tried like this. $$H(z,t)\leq \Phi(z,t)(S(z,t)-\rho_0)\leq\Psi(z_1,t_1)(S(z,t)-\rho_0)\leq\frac{S(z,t)-\rho_0}{S(z_1,t_1)-\rho_0}H(z_1,t_1).$$ From $0\leq \frac{1}{S- \rho_0 }[2H^2-\rho_0|A|^2H]$, I obtain $H\leq \frac{2n}{\rho_0}$ by Cauchy-Schwarz inequality. Thus, $$H(z,t)\leq\frac{S(z,t)}{2\rho_0-\rho_0}\frac{2n}{\rho_0}=\frac{2nS}{\rho_0^2}$$
Is this approach right? If so, how to continue? If not, how to explain?
Please, help me..
Thank you!