Is inverse matrix convex?

Assume $X, Y$ are positive definite and $\lambda \in [0,1]$. We know $Z(\lambda) = (1-\lambda) X + \lambda Y$ is positive definite and so does $Z^{-1}(\lambda)$. Let us use $(\ldots)'$ to represent derivative with respect to $\lambda$, we have:

$$ Z Z^{-1} = I_n \implies Z'Z^{-1} + Z ( Z^{-1} )' = 0_n \implies (Z^{-1})' = - Z^{-1} Z' Z^{-1} $$ Differentiate one more time and notice $Z'' = 0_n$, we get: $$(Z^{-1})'' = - (Z^{-1})' Z' Z^{-1} - Z^{-1}Z' (Z^{-1})' = 2 Z^{-1}Z' Z^{-1} Z' Z^{-1}\tag{*1}$$

Pick any random non-zero vector $u$ and consider following pair of vector/matrix valued functions:

$$v(\lambda) = Z'(\lambda) Z^{-1}(\lambda) u\quad\quad\text{ and }\quad\quad \varphi(\lambda) = u^T Z^{-1}(\lambda) u$$

$(*1)$ tell us

$$\varphi''(\lambda) = u^T (Z^{-1})''(\lambda) u = 2 v^T(\lambda) Z^{-1}(\lambda) v(\lambda) \ge 0\tag{*2}$$

because $Z^{-1}(\lambda)$ is positive definite. From this we can conclude $\varphi(\lambda)$ is a convex function for $\lambda$ over $[0,1]$. As a result, for any $\lambda \in (0,1)$, we have:

$$\begin{align}&(1-\lambda)\varphi(0) + \lambda\varphi(1) - \varphi(\lambda) \ge 0\\ \iff& u^T \left[ (1-\lambda) X^{-1} + \lambda Y^{-1} - ((1-\lambda) X + \lambda Y)^{-1}\right] u \ge 0\tag{*3} \end{align}$$ Since $u$ is arbitrary, this implies the matrix within the square bracket in $(*3)$ is positive semi-definite and hence:

$$(1-\lambda) X^{-1} + \lambda Y^{-1} \succeq ((1-\lambda) X + \lambda Y)^{-1}$$

Please note that when $Z' = Y - X$ is invertible, $v(\lambda)$ is non-zero for non-zero $u$. The inequalities in $(*2)$ and $(*3)$ become strict and the matrix within the square bracket in $(*3)$ is positive definite instead of positive semi-definite.

Let $\mathbf{P}=\mathbf{X}^{-1/2}\mathbf{Y}\mathbf{X}^{-1/2}$. By left- and right- multiplying both sides of the equation by $\mathbf{X}^{1/2}$, the inequality is equivalent to $(1-\lambda)I+\lambda\mathbf{P}^{-1} \succeq((1-\lambda)I+\lambda\mathbf{P})^{-1}$. Since $\mathbf P$ is positive definite, it can be unitarily diagonalised and hence we may assume that it is a diagonal matrix. So, the inequality reduces down to the scalar case $(1-\lambda)+\lambda p_{ii}^{-1} \ge ((1-\lambda)+\lambda p_{ii})^{-1}$, which is true because the function $f(t)=\frac1t$ is convex for $t>0$.