An 'obvious' inequality in Mean curvature flow My question is from the mean curvature flow paper by Gerhard Huisken. First $H:=g^{ij}h_{ij}=tr(h)$ where $h$ is the second fundamental form of the manifold. Then $A:=\left\{h_{ij}\right\}$.
It states that there is an obvious inequality by the Codazzi equation which is \begin{equation}
|\nabla H|^2\le n |\nabla A|^2
\end{equation} $n$ is the dimension of the manifold. So I try to expand everything out and make use of the operator norm. $$|\nabla H|^2=g^{lk}\nabla_l g^{ij}h_{ij}\nabla_kg^{pq}h_{pq}=g^{lk}g^{pq}g^{ij}\nabla_l h_{ij}\nabla_k h_{pq} $$ For the gradient of the second fundamental form $$|\nabla A|^2=g^{lk}g^{pq}g^{ij}\nabla_l h_{qi}\nabla_k h_{pj}\le ng^{pq}g^{ij}\nabla_l h_{qi}\nabla_k h_{pj}$$ I notice the inverse metric in $|\nabla H|$ represent a trace of $h_{ij}$ but the inverse metric in $|\nabla A|$ represent an inner product between two tensors. So  I don't know how to connect two magnitude.

Let $B_{ijk} = \nabla_iA_{jk}$. We want to maximize
$$
f(B) = \frac{g^{pq}B_{ipq}g_{jk}B^{ijk}}{B_{ijk}B^{ijk}} $$
First, note that it's obvious that the minimum value of $f$ is $0$. Therefore, the maximum occurs at a critical point of $f$.
Setting the directional derivative to $0$, we get, for any $2$-tensor $\dot{B}$ and some constant $c$,
\begin{align*}
0 &= g^{pq}g_{jk}B_{ipq}\dot{B}^{ijk} - c B_{ijk}\dot{B}^{ijk}\\
&= (g^{qr}g^{jk}B_{iqr}-cB_{ijk})\dot{B}^{ijk}.
\end{align*}
This implies that there is a vector $v$ such that
$$
B_{ijk} = v_ig_{jk}
$$
Therefore,
$$
g^{jk}B_{jk} = nv_i.
$$
The last two equations imply that
$$
f(B) = \frac{n^2|v|^2}{n|v|^2} = n.
$$
The Codazzi equations were not used here. If you take them into account, you might get an even better constant. In general, an inequality like this is called an improved Kato inequality.
