Here $\text{Rm}$ is the curvature tensor. When I try to compute
$$-\langle\nabla|\text{Rm}|^2,\nabla|\nabla \text{Rm}|^2\rangle\le4|\text{Rm}||\nabla \text{Rm}|^2|\nabla^2 \text{Rm}|,$$
I compute that: Firstly, I think
$$\text{Rm} =R_{ijkl}~\omega^i\otimes\omega^j\otimes\omega^k\otimes\omega^l.$$
Then $$\langle \text{Rm},\text{Rm}\rangle=g^{ia}g^{jb}g^{kc}g^{ld}R_{ijkl}R_{abcd}.$$
Choice normal coordinate, $\nabla g^{ij}=\nabla g_{ij}=0$. Then $$\nabla\langle \text{Rm},\text{Rm}\rangle=\nabla (g^{ia}g^{jb}g^{kc}g^{ld}R_{ijkl}R_{abcd}) =2g^{ia}g^{jb}g^{kc}g^{ld}R_{ijkl}\nabla R_{abcd}$$
But if we compute as above, it seem to be too complex that I can't finish it. Whether it's right? or there is some easy way?
The following is true for any tensor $B$:
$$\tag{1} |\nabla |B|^2| \le 2|B| |\nabla B| \ \ (\text{or } \big|\nabla |B|\big|\le |\nabla B|).$$
so we have
$$-\langle \nabla |\text{Rm}|^2, \nabla |\nabla \text{Rm}|^2 \rangle \le |\nabla |\text{Rm}|^2|\ |\nabla |\nabla \text{Rm}|^2|\le \big(2|\text{Rm} | |\nabla \text{Rm}|\big) \big( 2|\nabla \text{Rm}| |\nabla^2 \text{Rm}|\big),$$
which is what you need. $(1)$ can be proved by Cauchy schwarz inequality. For example, if $B = B_{i_1\cdots i_p}$, then under normal coordinates,
$$\begin{split} \left|\nabla |B|^2\right|^2 &= \sum_{\alpha} \left(\nabla_\alpha |B|^2\right)^2 \\ &= \sum_{\alpha} \left( \sum_{i_1, \cdots, i_p} \nabla_\alpha (B_{i_1\cdots i_p}^2)\right)^2 \\ &=4 \sum_{\alpha} \left( \sum_{i_1, \cdots, i_p} B_{i_1\cdots i_p} \nabla_\alpha B_{i_1\cdots i_p}\right)^2 \\ &\le 4 \sum_{\alpha} \left( \sum_{i_1, \cdots, i_p} (B_{i_1\cdots i_p} )^2\right)\left( \sum_{i_1, \cdots, i_p}(\nabla_\alpha B_{i_1\cdots i_p})^2 \right)\ \ \ \ \ \ \ \ \text{(Cauchy Schwarz's inequality)}\\ &= 4|B|^2 |\nabla B|^2\\ \Rightarrow |\nabla |B|^2| &\le 2|B| |\nabla B|. \end{split}$$