The geometric mean of two positive definite matrices $A, B$ is defined by $A\sharp B=A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2}$. It is known that the geometric mean is concave, i.e., $$\left(\sum_{i=1}^n A_i\right)\sharp \left(\sum_{i=1}^n B_i\right)\ge\sum_{i=1}^n A_i\sharp B_i$$ 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.

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.