Let $\rho(\cdot)$ be the spectral radius operator (NOT the spectral norm). I'm aware that $\rho(AB)\leq\rho(A)\rho(B)$ holds if matrices $A$ and $B$ commute. But what is wrong with the following series of inequalities: let $z$ be an arbitrary vector then $||ABz||\leq \rho(A)||Bz||\leq \rho(A)\rho(B)||z||$, thus maximum absolute eigenvalue of $AB$ (i.e. $\rho(AB)$) must be less than $\rho(A)\rho(B)$.
Please help me with this confusion,