I'm trying to show a quadratic equation of the form
$$\frac{1}{2} a' \Sigma^{-1} a x^2 - a' \Sigma^{-1} \iota x + \frac{1}{2} \iota' \Sigma^{-1} \iota$$ has a real solution. Here $\Sigma$ is an $N \times N$ variance-covariance matrix (so both itself and its inverse are positive-definite), and $\iota = (1,...,1)'$ is an $N \times 1$ vector of ones. When plugging the coefficients in the quadratic formula, I need to show that
$$(a' \Sigma^{-1} \iota)^2 \geq ( a' \Sigma^{-1}a) \cdot (\iota' \Sigma^{-1} \iota)$$
This seems like a form of Cauchy-Schwartz to me, but I can't seem to prove it.
Indeed your issue cannot receive a positive answer. Here is why.
$$\begin{pmatrix}ax \ \ \iota\end{pmatrix}^T \underbrace{\begin{pmatrix}\ \ \Sigma^{-1}&-\Sigma^{-1}\\ -\Sigma^{-1}& \ \ \Sigma^{-1}\end{pmatrix}}_M \begin{pmatrix}ax\\ \iota\end{pmatrix}\tag{1}$$
$\operatorname{rank}(M)\le N$ because the $N$ last columns of $M$ are the opposite of its first $N$ columns. Therefore $\operatorname{rank}(M)=N$ because the upper left $N \times N$ block of $M$ has rank $N$ exactly.
It means that $M$ can be diagonalized in this way:
$$M=P^T diag(\lambda_1, \cdots, \lambda_N, 0, 0, \cdots 0) P$$
In this way, you will be able to write (1) under the form:
$$\begin{pmatrix}ax \ \ \iota\end{pmatrix}^T P^T diag(\lambda_1, \cdots, \lambda_N, 0, 0, \cdots 0) P \begin{pmatrix}ax\\ \iota\end{pmatrix}=0\tag{2}$$
for a certain $2N \times 2N$ matrix $P$.
which is possible because (2) can be reduced to an expression involving only the $N$ first entries of the vectors and the upper left blocks of the matrices :
$$\begin{pmatrix}ax\end{pmatrix}^T Q^T diag(\lambda_1, \cdots, \lambda_N) Q \begin{pmatrix}ax\end{pmatrix}=0\tag{3}$$
where $Q$ is the upperleft $N \times N$ block of $P$.
Said otherwise:
$$(Q ax)^T diag(\lambda_1, \cdots, \lambda_N) (Q ax)=0\tag{4}$$
but factoring out $x$, we remain with a strictly positive factor due to the fundamental property of a positive definite matrix.