This problem comes arises from the Markowitz Portfolio optimization with $n$ risky assets.
Suppose $\mathbf{1} = (1, 1, ..., 1)\in\mathbb{R}^{n}$ and let $\mathbf{z}\in\mathbb{R}^{n}$ be the vector of expected returns and $\Sigma$ be the covariance matrix for the assets. I have the matrix $$D = \begin{bmatrix}A & B \\ B & C\end{bmatrix}$$ where $$ \begin{align*} A & = \mathbf{1}^{\top}\Sigma^{-1}\mathbf{1} \\ B & = \mathbf{1}^{\top}\Sigma^{-1}\mathbf{z} \\ C & = \mathbf{z}^{\top}\Sigma^{-1}\mathbf{z}. \end{align*} $$ Question: How can I prove that the determinant of $D$ is positive?
My attempt: I know $A,B,C$ are all positive because $\Sigma^{-1}$ is positive definite. Also, $D$ is symmetric, but I do not think I have enough to conclude $D$ is positive definite. I tried finding the eigenvalues, but arrived that the expression $$\lambda = \frac{A+C\pm\sqrt{(A+C)^{2} - 4(AC - B^{2})}}{2},$$ from which I could not get any further.
I found the answer thanks to Michal's hint. Since $\Sigma^{-1}$ is positive definite, $\langle u, v\rangle =u^{\top}\Sigma^{-1}v$ defines an inner product on $\mathbb{R}^{n}\times\mathbb{R}^{n}$. By the Cauchy Schwarz inequality,
$$|\langle 1, z\rangle|^2 = B^{2}\leq |\langle 1, 1\rangle||\langle z, z\rangle|=AC$$
Hence, $$\Delta = AC-B^2\geq 0$$