For any square matrix $C$ with real entries, denote by $\rho(C)$ its spectral radius, i.e. the maximum magnitude of its eigenvalues. For symmetric matrices $A$ and $B$ with $AB=BA$ show that $$\rho(AB)\le \rho(A)\rho(B)$$
I think simultaneous diagonalization of $A$ and $B$ is to be used here, but couldn't find my way out.
Also will the proposition hold if the condition of symmetry is dropped?