On closed smooth manifold $M$ $$ \partial_t H \ge \Delta H + \frac{1}{n} H^3 $$ Let $\varphi$ be the solution of $$ \frac{d \varphi}{dt}=\frac{1}{n}\varphi ^3 ~~~~ \\ \varphi(0)=H_\min(0) >0 $$ Then , we get $$ \partial_t(H-\varphi) \ge \Delta(H-\varphi) + \frac{1}{n} (H^3-\varphi^3) $$ Then, how to use maximum principle to get $$ H\ge \varphi ~~? $$ I use the maximum principle in Evans' book as picture below, but in above the $c$ is not $\ge 0$.
The question is from 252 page of Huisken, Gerhard, Flow by mean curvature of convex surfaces into spheres, J. Differ. Geom. 20, 237-266 (1984). ZBL0556.53001. as picture below
You are not really using Evan's results since it considers bounded open sets, which has boundary $\Gamma_T$. Indeed just prove directly:
Proof: Let $\epsilon >0$ and consider $f_\epsilon = f+ \epsilon t$. Then $$ (\partial_t - \Delta ) f_\epsilon \ge \epsilon, \ \ \ f_\epsilon \ge 0.$$ If $f_\epsilon <0$ at some points, since $M\times [0,T]$ is compact, let $X=(x, t)$ be where $f_\epsilon$ is minimized. Then $t>0$ and so at this point, $$ \partial_t f_\epsilon \le 0, \Delta f_\epsilon \ge 0$$ (We use also that $M$ has no boundary to conclude the second inequality). Thus $(\partial_t -\Delta )f_\epsilon \le 0$, which is impossible. Thus $f_\epsilon \ge 0$ and so $f\ge -\epsilon T$. Then $f\ge 0$ by taking $\epsilon \to 0$.
Now note that both $H, \phi\ge H_{min}(0)>0$. Thus $g = H-\phi$ satisfies
$$(\partial_t -\Delta )g \ge \frac{1}{n}g (H^2 + H\phi + \phi^2) \ge Cg$$
for $C= \frac{3}{n}H^2_{min}(0)$. Then $f = e^{-Ct}g$ satisfies $$(\partial_t -\Delta)f \ge 0$$ and $f\ge 0$ at $t=0$. So the weak maximum principle implies $f\ge 0$ for all time $t\in [0,T]$ and so $H\ge \phi$.
I assume this theorem can be found in any books on Ricci flows.