How to show show $\langle \nabla H , \nabla |A|^2 \rangle=2H|\nabla H|^2$

Question

Consider a embedded submanifold. Let $H=g^{ij}h_{ij}$ is mean curvature, $|A|^2=g^{ij}g^{kl}h_{ik}h_{jl}$. $h_{ij}$ is the second fundamental form and $g_{ij}$ is induced metric. For show $\langle \nabla H , \nabla |A|^2 \rangle=2H|\nabla H|^2$. First, choice a normal coordinate , then $$ \langle \nabla H , \nabla |A|^2 \rangle=\nabla_i h_{jj}\nabla_i(h_{aa}h_{bb}) \\ ~~~~~~~~~~~~~~~~~~~~=2h_{bb}\nabla_i h_{jj}\nabla_i h_{aa} \\ ~~~~~~~~~=2H|\nabla H|^2 $$ But I need it in integral, so , the normal coordinate can't be used. So, $$ \langle \nabla H , \nabla |A|^2 \rangle= g^{kl}\nabla_k(g^{ij}h_{ij})\nabla_l(g^{ab}g^{cd}h_{ac}h_{bd}) \\ ~~~~~~~~~~~~~~~~~=2g^{kl}g^{ij}g^{ab}g^{cd}h_{ac}\nabla_k h_{ij}\nabla_l h_{bd} \\ ~~~~~~~~~~~=2g^{kl}g^{ab}g^{cd}h_{ac}\nabla_kH \nabla_l h_{bd} $$ But I don't know how to deal $\nabla_l h_{bd}$.