Let $x = (0,0,...,0,1)^T$ and $S,R$ symmetric positive definite matrices. I want to see if the following is true:
$\frac{1}{2} x^ T\left(R + S \right)x \geq x^T \left(S^{\frac{1}{2}} \left(S^{-\frac{1}{2}} \,R \, S^{-\frac{1}{2}} \right)^{\frac{1}{2}} S^{\frac{1}{2}} \right)x$
Aditional info: It is to see if the quadratic form with this particular vector is g-concave under the geodesic $\gamma(z) = S^{\frac{1}{2}} \left(S^{-\frac{1}{2}} \,R \, S^{-\frac{1}{2}} \right)^{z} S^{\frac{1}{2}}$ induced by the Riemannian metric $g_\Sigma (A,B) = Tr(\Sigma^{-1} A \Sigma^{-1} B)$ on the manifold of SPD matrices. So far I don't see how to prove or disprove this. Any help would be appreciated.
Yes, it is true, because $$ \frac12(R+S)-S^{1/2}\left(S^{-1/2}RS^{-1/2}\right)^{1/2}S^{1/2} =\frac12S^{1/2}\left[\left(S^{-1/2}RS^{-1/2}\right)^{1/2}-I\right]^2S^{1/2} \ge0. $$