This question is very similar to this one.
Let $B$ be a positive semi-definite matrix and $B = \begin{bmatrix} B_{11} & B_{12} \\ B_{12}' & B_{22} \end{bmatrix}$ where $B_{11}$ is $p \times p$. then $\lambda_1(B) \le \lambda_1(B_{11}) + \lambda_1(B_{22})$ where $\lambda_1$ is the largest eigenvalue of the matrix in argument.
How can I show it using extremal representation of the maximum eigenvalue of a symmetric matrix.
$ \lambda_1(B) = max_{||x||=1} x'Bx \\ = max_{||x||=1} [x_1' x_2']\begin{bmatrix} B_{11} & B_{12} \\ B_{12}' & B_{22}\end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \\ = max_{||x||=1} [x_1' x_2']\begin{bmatrix} B_{11}x_1 + B_{12}x_2 \\ B_{12}'x_1 + B_{22}x_2\end{bmatrix} \\ = max_{||x||=1} { x_1'B_{11}x_1 + x_1'B_{12}x_2 + x_2'B_{12}'x_1 + x_2'B_{22}x_2 } $
If $x_1'B_{12}x_2 + x_2'B_{12}'x_1 $ can go to zero then I am done. But this is certainly not the case.
Where am I making a mistake?