How to prove the matrix fractional function is convex by definition It is well known that the matrix fractional function $f(\mathbf{w},\boldsymbol{\Omega})=\mathbf{w}^T\boldsymbol{\Omega}^{-1}\mathbf{w}$ is jointly convex with respect to $\mathbf{w}$ and $\boldsymbol{\Omega}$ for $\mathbf{w}\in\mathbb{R}^n$ and $\boldsymbol{\Omega}\in\mathbb{S}_+^{n\times n}$, where $\mathbb{S}_+^{n\times n}$ denotes the set of all $n\times n$ positive definite matrix. 
Now I want to prove it based on the definition of convex functions. That is, I need to prove that the following inequality holds for any $\mathbf{w}_1,\mathbf{w}_2\in\mathbb{R}^n$, $\boldsymbol{\Omega}_1,\boldsymbol{\Omega}_2\in\mathbb{S}^{n\times n}_+$, and $\alpha\in[0,1]$:
$$\alpha\mathbf{w}_1^T\boldsymbol{\Omega}_1^{-1}\mathbf{w}_1+\beta\mathbf{w}_2^T\boldsymbol{\Omega}_2^{-1}\mathbf{w}_2-(\alpha\mathbf{w}_1+\beta\mathbf{w}_2)^T(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}(\alpha\mathbf{w}_1+\beta\mathbf{w}_2)\ge 0,$$
where $\beta=1-\alpha$. In order to prove this inequality, I first simplify the left-hand side as
\begin{align*}
&\alpha\mathbf{w}_1^T\boldsymbol{\Omega}_1^{-1}\mathbf{w}_1+\beta\mathbf{w}_2^T\boldsymbol{\Omega}_2^{-1}\mathbf{w}_2-(\alpha\mathbf{w}_1+\beta\mathbf{w}_2)^T(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}(\alpha\mathbf{w}_1+\beta\mathbf{w}_2)\\
=&(\alpha\mathbf{w}_1^T\boldsymbol{\Omega}_1^{-1}\mathbf{w}_1-\alpha^2\mathbf{w}_1^T(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_1)+(\beta\mathbf{w}_2^T\boldsymbol{\Omega}_2^{-1}\mathbf{w}_2-\beta^2\mathbf{w}_2^T(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_2)\\
&-2\alpha\beta\mathbf{w}_1^T(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_2\\
=&\alpha\beta\mathbf{w}_1^T\boldsymbol{\Omega}_1^{-1}\boldsymbol{\Omega}_2(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_1+\alpha\beta\mathbf{w}_2^T\boldsymbol{\Omega}_2^{-1}\boldsymbol{\Omega}_1(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_2\\
&-2\alpha\beta\mathbf{w}_1^T(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_2\\
=&\alpha\beta\mathbf{w}_1^T(\alpha\boldsymbol{\Omega}_1\boldsymbol{\Omega}_2^{-1}\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_1)^{-1}\mathbf{w}_1+\alpha\beta\mathbf{w}_2^T(\alpha\boldsymbol{\Omega}_2+\beta\boldsymbol{\Omega}_2\boldsymbol{\Omega}_1^{-1}\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_2\\
&-2\alpha\beta\mathbf{w}_1^T(\alpha\boldsymbol{\Omega}_1+\beta\boldsymbol{\Omega}_2)^{-1}\mathbf{w}_2
\end{align*}
I cannot prove the inequality based on this simplification. Is there any way to prove the inequality? Thanks.
 A: It suffices to establish midpoint convexity. So we show that
\begin{equation*}
f(\tfrac{w+v}{2},\tfrac{A+B}{2}) \le \tfrac12 f(w, A) + \tfrac12 f(v,B).
\end{equation*}
In other words, we show that
\begin{equation*}
\left\langle\tfrac{w+v}{2},\left(\tfrac{A+B}{2}\right)^{-1}\tfrac{w+v}{2}\right\rangle \le \tfrac12 f(w, A) + \tfrac12 f(v,B),
\end{equation*}
which simplifies to showing that
\begin{equation*}
\tag{*}
w^TA^{-1}w+v^TB^{-1}v \ge (w+v)^T(A+B)^{-1}(w+v).
\end{equation*}
This follows easily from recalling that a matrix with positive definite diagonal blocks is positive definite if and only if its Schur complement is psd. Thus, it follows that since
\begin{equation*}
\begin{bmatrix}
w^TA^{-1}w & w^T\\
w & A
\end{bmatrix} \succeq 0,\ \text{and}\ 
\begin{bmatrix}
v^TB^{-1}v & v^T\\
v & B
\end{bmatrix} \succeq 0,\ \implies
\begin{bmatrix}
w^TA^{-1}w +v^TB^{-1}v & w^T+v^T\\
w+v & A+B
\end{bmatrix} \succeq 0.
\end{equation*}
Taking Schur complements of the final matrix above we obtain (*) as desired.
