Comparison of the largest eigenvalues of two symmetric matrices Let $$A=\left(\begin{array}{ccccc|cccc}0&\cdots&0&0&0&0&0&a\\\vdots&\cdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\0&\cdots&0&0&0&0&0&a\\0&\cdots&0&0&b&0&0&0\\0&\cdots&0&b&0&b&0&0\\\hline 0&\cdots&0&0&b&0&\frac{1}{\sqrt{2}}&0\\0&\cdots&0&0&0&\frac{1}{\sqrt{2}}&0&\frac{1}{\sqrt{2}}\\a&\cdots&a&0&0&0&\frac{1}{\sqrt{2}}&0
\end{array}\right),$$ be an $n\times n$ matrix, where $a=\sqrt{\frac{6}{7}}$ and $b=\frac{1}{\sqrt{2}}$.
Let $B$ be same as $A$ except the second diagonal block which is $\left(\begin{array}{ccc}0&\sqrt{\frac{2}{3}}&\sqrt{\frac{8}{21}}\\\sqrt{\frac{2}{3}}&0&0\\\sqrt{\frac{8}{21}}&0&0\end{array}\right).$ With help of matlab I have verified that largest eigenvalue of $A$ is greater than that of $B$. How to prove it? I thought of doing it with calculating determinant of partitioned matrices but couldn't succeed.

Not an answer, just too long for a comment.
Well, I decided to try and brute-force the computation of the characteristic polynimial for $A$ and got the following:
let $D_n$ be the determinannt of $tI_n-A$ (asumme $n\geq6$ so $A$ will contain the required $a$'s), so:
$$D_n = tD_{n-1} -a^2t^{n-6}(t^4-6b^2t^2+b^4)$$
and:
$$D_6 = t^6+(b-a^2)t^4 -3b^2t^3+(b^4-2b^3)t^2+(3a^2b^2-ab^4)t-a^2b^4$$
