Product between normal and hyponormal operators which commute is hyponormal

Question

Let $A\in \mathcal{L}(E)$ be a normal operator i.e $A^{*}A=AA^{*}$.

Let $B\in \mathcal{L}(E)$ be an hyponormal operator i.e. $B^*B\geq BB^*$. If $AB=BA$. Why $AB$ is hyonormal?

I try to apply the following theorem:

Fuglede's theorem: Let $T,S\in \mathcal{L}(E)$. If $T$ is normal and $TS=ST$, then $TS^*=S^*T$.

Answer 1

Since $A$ is normal and $AB=BA,$ we get $$AB^*=B^*A.$$ Similarly, since $A^*$ is normal, and $A^*B^*=B^*A^*,$ we get $$A^*B=BA^*.$$

Now note that

$(AB)^*AB=B^*(A^*A)B=B^*(AA^*)B=(B^*A)(A^*B)=(AB^*)(BA^*)=A(B^*B)A^*.$

Hence $$A(B^*B)A^*\geq A(BB^*)A^*=(AB)(AB)^*.$$

This completes the proof.