Computing the corresponding matrices It can be seen that, in the $2\times2$ case, $\rho(G_{jacobi})^2=\rho(G_{G-S})$. This implies that the answer to your question is no.
Yes. From pg. 233 of A Friendly Introduction to Numerical Analysis by Brian Bradie, we have the following:
Theorem. Suppose $A$ is a an $n \times n$ matrix. If $ a_{ii} > 0$ for each $i$ and $a_{ij} \leq 0$ whenever $ i \neq j$, then one and only one of the following statements holds:
$0 \leq \rho (T_{gs}) < \rho (T_{jac}) < 1$
$1 < \rho (T_{jac}) < \rho (T_{gs})$
$\rho (T_{jac}) = \rho (T_{gs}) = 0$
$\rho (T_{jac}) =\rho (T_{gs}) = 1$
where $T$ is the iteration matrix that arises for each method, and $\rho (T)$ denotes the spectral radius of $T$.
Thus, for this larger class of matrices, the methods converge and diverge together. When they converge, Gauss-Seidel converges faster; when they diverge, Gauss-Seidel again does so faster.
If you want the proof of this, Bradie cites the following sources:
A. Ralston and P. Rabinowitz, A First Course in Numerical Analysis, 2nd edition, McGraw-Hill, New York, 1978.
D. M. Young, Iterative Solution of Large Linear Systems, Academic Press, New York, 1971.
Best Answer
For iterational methods in the canonical form $$ \mathbf P(\mathbf x_{n+1} - \mathbf x_n) = \mathbf f - \mathbf A \mathbf x_n $$ the nessesary and sufficient condition is given by roots of $$ \det (\mathbf A - \mathbf P + \lambda \mathbf P) = 0 \tag{*} $$ which all should satisfy $|\lambda| < 1$. It can be easily shown that those $\lambda$ are simply eigenvalues of $\mathbf B$, where $\mathbf B$ is iteration matrix $$ \mathbf B = \mathbf P^{-1}(\mathbf P - \mathbf A). $$
Jacobi case. The matrix $\mathbf P$ for the Jacobi method is given by $$ \mathbf P = \operatorname{diag}(\mathbf A). $$ The equation $(*)$ gives $$ \det (\mathbf A - \mathbf P + \lambda \mathbf P) = \begin{vmatrix} 2\lambda & 1 & -1\\ -2 & 2\lambda & -2\\ 1 & 1 & 2\lambda \end{vmatrix} = 8\lambda^3 + 10\lambda = 0. $$ The roots are $\lambda_1 = 0, \lambda_{2,3} = \pm\sqrt\frac{5}{4}i$. Obviously, $|\lambda_{2,3}| = 5/4 > 1$. The method does not converge.
Seidel case. The matrix $\mathbf P$ for the Seidel method is given by $$ \mathbf P = \operatorname{diag}(\mathbf A) + \operatorname{lower}(\mathbf A). $$ The equation $(*)$ gives $$ \det (\mathbf A - \mathbf P + \lambda \mathbf P) = \begin{vmatrix} 2\lambda & 1 & -1\\ -2\lambda & 2\lambda & -2\\ \lambda & \lambda & 2\lambda \end{vmatrix} = 2\lambda(4\lambda^2 + 6\lambda - 1) = 0. $$ The roots are $\lambda_1 = 0, \lambda_{2,3} = \frac{-3 \pm \sqrt{13}}{4} \approx {-1.65139, 0.151388}$. The eigenvalue $\lambda_3$ violates the $|\lambda|<1$ condition, thus the method wont converge. Numerical experiment shows the same.