Given matrices
$$A = \begin{bmatrix} 0 & 0 & 0 & 1/3 \\ 0 & 0 & 0 & 1/2 \\ 0 & 0 & 0 & 1 \\ 1/3 & 1/2 & 1 & 0 \end{bmatrix}$$
and
$$B = \begin{bmatrix} 0 & 0 & 1/2 & 1/3 \\ 0 & 0 & 1 & 1/2 \\ 1/2 & 1 & 0 & 0 \\ 1/3 & 1/2 & 0 & 0 \end{bmatrix}$$
show that $\| B \|_2 > \| A \|_2$ without actually calculating the largest (in absolute value) eigenvalues.
Since for symmetric matrices:
$$\| M \|_2 = \max_{\|x\|_2 = 1} |x^{t}Mx|$$
I've tried to show that the set $|x^{t}Ax|$ is dominated by the set $|x^{t}Bx|$, but without success. I have also tried to relate to the Frobenius norm since the Frobenius norm of $B$ is clearly bigger than that of $A$, but I can't make the inequalities between Frobenius and spectral norm work properly for this. In fact, I have calculated the spectral norms of both matrices: for $A$ it is 1.1667 and for $B$ it is 1.2676. The closeness between these values also suggests a finer argument is needed than just converting to the Frobenius norm.
Note: I want to do this problem without actually calculating the norms because this is only the first instance of a class of similar problems for larger matrix sizes. The question is from trying to understand the corresponding claim, made as though obvious, in 0509153 (Example: Ordered Searching 2).