If $M_t$ evolve under volume preserving mean curvature flow, assume $$ A=(h_{ij}) $$ is the second fundamental form of $M_t$, and $$ |A|^2=h_i^jh_j^i $$ then, we have $$ \partial_t|A|^2= \Delta |A|^2- 2|\nabla A|^2+ 2|A|^4 - 2h ~tr(A^3) \tag{1} $$ where $tr(A^3) = h_i^j h_j^k h_k^i$, and $$ h(t)=\frac{\int_{M_t} HdS }{\int_{M_t} dS} $$ Now, how to get $$ \partial _t |A|\le \Delta |A| + |A|^3 + h|A|^2 $$ from (1) ?
PS: this question origins from the 337th page of
Li, Haozhao, The volume-preserving mean curvature flow in Euclidean space, Pac. J. Math. 243, No. 2, 331-355 (2009). ZBL1182.53061.
Note that
$$\partial _t |A|^2 = 2|A|\; \partial _t |A|$$ and
$$\Delta |A|^2 = 2|A| \Delta |A| + 2|\nabla |A||^2.$$
Using Kato's inequality we have $|\nabla |A|| \le |\nabla A|$ and
$$ |\operatorname{tr} (A^3)| \le |A|^3$$
by applying Cauchy-Schwarz twice, we obtain
$$2|A| \partial _t |A| \le 2|A| \Delta |A| + 2|A|^4 + 2h |A|^3, $$ which implies the inequality.