The geometric mean of two positive definite matrices $A, B$ is defined by $A\sharp B=A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}$. The following inequality holds true $$\left(\sum\_{i=1}^n A\_i\right)\sharp \left(\sum\_{i=1}^n B\_i\right)\ge\sum\_{i=1}^n A\_i\sharp B\_i$$ for positive definite matrices $A\_i, B\_i$, $i=1\ldots, n$. 


The spectral mean is defined by Fiedler and Ptak as 
$A\natural B=(A^{-1}\sharp B)^{1/2}A(A^{-1}\sharp B)^{1/2}$. Is the spectral mean also concave? That is, whether 


$$\left(\sum\_{i=1}^n A\_i\right)\natural \left(\sum\_{i=1}^n B\_i\right)\ge\sum\_{i=1}^n A\_i\natural B\_i$$ for positive definite matrices $A\_i, B\_i$, $i=1\ldots, n$? 


The inequality here is Loewner order. 



Concavity of Spectral mean
Some experiments reveal that the said inequality **does not** hold for the spectral mean.


Here is a random (i.e., mindless) counterexample:


\begin{equation\*}
 A\_1=\begin{bmatrix}
 29 & 15\\\\
 15 & 26
\end{bmatrix},\quad A\_2=\begin{bmatrix}
5 & 0\\\\
0 & 5
\end{bmatrix},\quad B\_1=\begin{bmatrix}
4 & -16\\\\
-16 & 113
\end{bmatrix},\quad B\_2=\begin{bmatrix}
18 & -12\\\\
-12 & 16
\end{bmatrix}.
\end{equation\*}


With this choice, we have


\begin{equation\*}
 (A\_1+A\_2)\natural (B\_1+B\_2) = \begin{bmatrix}
 22.3606 & 0.4475\\\\
 0.4475 & 58.3661
\end{bmatrix},\quad (A\_1\natural B\_1)+(A\_2\natural B\_2) =\begin{bmatrix}
15.2404 & -2.2975\\\\
-2.2975 & 58.5164
\end{bmatrix}.
\end{equation\*}


A quick calculation shows that the difference between the two matrices is an indefinite matrix, so the alleged inequality does not hold.


