Prove $\frac 12(AB+BA)$ is hermitian when $A$ and $B$ are hermitian

Question

$A$ and $B$ are hermitian, so I know they must commute. So $AB-BA = 0$. But I don't think I can get very far with that. I just totally don't know how to start. Some help would be very appreciated.

Answer 1

Note that $$(AB)^H=B^HA^H=BA\\(BA)^H=A^HB^H=AB$$therefore$${1\over 2}(AB+BA)^H={1\over 2}((AB)^H+(BA)^H)={1\over 2}(AB+BA)$$