Let $M$ be a positive definite matrix. Can one show that $\sqrt{x^\top M^{-1}x}$ is convex in M?
In other words, for any $\lambda\in[0,1]$, $x\in\mathbb{R}^d$, $A,B\in\mathbb{R}^{d\times d}$ positive definite, is the following true: $$ (1-\lambda)\sqrt{x^\top (A)^{-1}x}+\lambda \sqrt{x^\top B^{-1}x} \geq \sqrt{x^\top ((1-\lambda)A+\lambda B)^{-1}x} $$
It is known that without the square-root, $x^\top M^{-1}x$ is convex in M. This is by showing that the second derivative is positive in any direction, see e.g. this short note. I couldn't quite replicate the same approach on $\sqrt{x^\top M^{-1}x}$, since the additional square-root makes the second derivative expression much more complicated, and it does not easily reduce to a quadratic form.
Any help is appreciated!
Edit:
Here is what I have tried. Denote $D = (1-\lambda)A+\lambda B$ and $C = B-A$. We will try to show that the following function is convex in $\lambda$ for any $x,A,B$ and $\lambda\in[0,1]$: $$ f(\lambda) = \sqrt{x^\top ((1-\lambda)A+\lambda B)^{-1}x} $$
We do this by proving that $f''\geq 0$ for all $x,A,B$ and $\lambda\in[0,1]$.
According to my calculations: $$ f'' = -\frac{1}{4}\frac{(x^\top D^{-1}CD^{-1} x)^2}{(x^\top D^{-1}x)^{1.5}} + \frac{x^\top D^{-1}CD^{-1}CD^{-1} x}{(x^\top D^{-1}x)^{0.5}} $$
If the following is true: $$ (x^\top D^{-1}CD^{-1}CD^{-1} x)(x^\top D^{-1}x) \geq (x^\top D^{-1}CD^{-1} x)^2 $$
This would imply that $$ f'' \geq \frac{3}{4}\frac{(x^\top D^{-1}CD^{-1} x)^2}{(x^\top D^{-1}x)^{1.5}}\geq 0 $$ since D is p.d.
Now, it is left to show $(x^\top D^{-1}CD^{-1}CD^{-1} x)(x^\top D^{-1}x) \geq (x^\top D^{-1}CD^{-1} x)^2$. I've verified this with simulation and it seems to be correct, but I wasn't able to prove it.