Let $H$ denote the Hamiltonian operator and let $|\psi\rangle$ be an eigenstate of $H$ with corresponding eigenvalue $E$, i.e., $$H|\psi\rangle=E|\psi\rangle$$
I want to show that $$\langle\psi |[H,A]|\psi\rangle=0$$ for any operator $A$, where $[H,A]$ denotes the commutator of $H$ and $A$.
I've started as follows: \begin{align}\langle\psi |[H,A]|\psi\rangle & = \langle\psi|HA-AH|\psi\rangle \\ & = \langle\psi|HA|\psi\rangle -\langle \psi|AH|\psi\rangle \\ & = \langle\psi|HA|\psi\rangle-E\langle \psi|A|\psi\rangle\end{align}
I'm not sure how to deal with the $\langle \psi|HA|\psi\rangle $ term though.
Taking the adjoint of $$ H\vert \psi\rangle = E\vert\psi\rangle$$ gives $$\langle \psi\vert H = \langle \psi\vert E $$ since $H$ is Hermitian and $E$ is real. So you have $$ \langle\psi\vert HA\vert \psi\rangle = E\langle\psi\vert A\vert \psi\rangle $$