Let $\mathbf A=\left[\begin{matrix}\mathbf A\_{11}&\mathbf A\_{12}\\ \mathbf A\_{21}&\mathbf A\_{22}\end{matrix}\right]$ be a positive semi-definite matrix, $\mathbf A\_{ij}\in\mathbb C^{n\times n}$ and $rank(\mathbf A)=n$.


**Prove or disprove** that
$$
\lambda\_{max}(\mathbf A) + \lambda\_{max}(\mathbf A\_{11}+\mathbf A\_{22}) \leq
\lambda\_{max}\left(\left[\begin{matrix}tr(\mathbf A\_{11})&tr(\mathbf A\_{12})\\ tr(\mathbf A\_{21})&tr(\mathbf A\_{22})\end{matrix}\right]\right) + tr(\mathbf A\_{11}+\mathbf A\_{22}),
$$
where $\lambda\_{max}$ denotes the largest eigenvalue and $tr$ denotes the trace.


I tried small values of $n$. For $n=1$ we have "$=$" (the same terms in the LHS and RHS). I have a rather long proof for $n=2$ (in my proof I also replaced $\mathbb C$ by $\mathbb R$). 


It seems that some results on [(completely) positive maps](https://en.wikipedia.org/wiki/Completely_positive_map)
are somehow related to the inequality ( indeed, all the maps
$$
\mathbf A\rightarrow \mathbf A\_{11}+\mathbf A\_{22},\qquad 
\mathbf A\rightarrow \left[\begin{matrix}tr(\mathbf A\_{11})&tr(\mathbf A\_{12})\\ tr(\mathbf A\_{21})&tr(\mathbf A\_{22})\end{matrix}\right],\qquad
\mathbf A\rightarrow 
tr(\mathbf A\_{11}+\mathbf A\_{22})
$$
are positive). By Theorem 2.3.7 (The Russo-Dye Theorem) from the book R. Bhatia, Positive definite matrices,
$$
\lambda\_{max}\left(\left[\begin{matrix}tr(\mathbf A\_{11})&tr(\mathbf A\_{12})\\ tr(\mathbf A\_{21})&tr(\mathbf A\_{22})\end{matrix}\right]\right)\leq n\lambda\_{max}(\mathbf A),
$$
which is useless since we need inequality of the type $\lambda\_{max}(\mathbf A)\leq\dots$



Upper bound of the largest eigenvalue of a PSD block matrix in terms of blocks
Write $A=\sum w\_i\otimes w\_i$ where $w\_i$ are orthogonal and $\|w\_1\|=\max\_i\|w\_i\|$. Also write $w\_i=(u\_i,v\_i)$. Then the inequality in question is just 
$$
\|u\_1\|^2+\|v\_1\|^2+\max\_{e:\|e\|=1}\sum\_i[|(u\_i,e)|^2+|(v\_i,e)|^2]
\\
\le \left\|\begin{pmatrix}\sum\_i\|u\_i\|^2&\sum\_i(u\_i,v\_i)\\\sum\_i(v\_i,u\_i)&\sum\_i\|v\_i\|^2\end{pmatrix}\right\|+\sum\_i(\|u\_i\|^2+\|v\_i\|^2)
$$
Now, cancel the first term with the corresponding term on the RHS and estimate
$
\sum\_{i\ge 2}[|(u\_i,e)|^2+|(v\_i,e)|^2]
$
by 
$
\sum\_{i\ge 2}(\|u\_i\|^2+\|v\_i\|^2)
$.
Also, note that the norm of the sum of several positive definite matrices is not less than the norm of each of them, so we can remove all $i\ge 2$ in the matrix and reduce the problem to showing that
$$
|(u\_1,e)|^2+|(v\_1,e)|^2\le \left\|\begin{pmatrix}\|u\_1\|^2&(u\_1,v\_1)\\(v\_1,u\_1)&\|v\_1\|^2\end{pmatrix}\right\| 
$$
However, we can write the matrix on the right as the sum of positive definite matrices of the kind $\begin{pmatrix}|(u\_1,e\_j)|^2& (u\_1,e\_j)(e\_j,v\_1)\\(v\_1,e\_j)(e\_j,u\_1)&|(v\_1,e\_j)|^2\end{pmatrix}$ where $e\_j$ is an orthonormal basis with $e\_1=e$. Again, we can remove all matrices with $j\ge 2$, after which we get an identity.


