Maximum principle in a proof of mean curvature flow 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



 A: You are not really using Evan's results since it considers bounded open sets, which has boundary $\Gamma_T$. Indeed just prove directly: 

Weak maximum principle: Let $f : M\times [0,T]\to \mathbb R$ satisfies 
  $$ (\partial_t - \Delta) f \ge 0.$$ 
  If $f\ge 0$ at $t=0$, then $f\ge 0$ on $[0,T]$. 

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. 
