Is $\rho(AB) \leq \rho(A)\rho(B)$ if $A, B$ are forward & backward Gauss-Siedel iteration matrix?

Question

If I understand correctly, to solve $Ax = b$, if we denote $A = D + L + U$, then the iteration matrices of forward & backward Gauss-Seidel method are $G_{GS} = -(D + L)^{-1}U$ and $G_{SG} = -(D + U)^{-1}L$ respectively. Now that I can use the fact that $G_{GS}, G_{SG} < 1$, how can I show $G_{GSG}$, the iteration matrix of symmetric Gauss-Seidel method also converges? Can I say $\rho(G_{GSG} = G_{GS} \times G_{SG}) < \rho(G_{GS}) \rho(G_{SG}) < 1$? If yes, how does the inequality stand? If not, how can I show $\rho(G_{GSG}) < 1$? Thanks in advance.