Concavity of Spectral mean

Question

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.

Answer 1

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.