[proofplan] The proof rewrites the numerator as an inner product for the positive definite quadratic form induced by $\Sigma$. A direct Cauchy--Schwarz argument for this inner product gives the upper bound $J(a) \leq d^\top \Sigma^{-1}d$. The equality condition in that Cauchy--Schwarz inequality identifies exactly when $a$ is a non-zero scalar multiple of $\Sigma^{-1}d$, which proves both the maximisers and the maximum value. [/proofplan]

[step:Define the inner product generated by $\Sigma$ and rewrite the criterion] Since $\Sigma$ is real symmetric positive definite, the map \begin{align*} (\cdot,\cdot)_\Sigma: \mathbb{R}^p \times \mathbb{R}^p &\to \mathbb{R} \\ (x,y) &\mapsto x^\top \Sigma y \end{align*} is an inner product on $\mathbb{R}^p$. Define \begin{align*} b := \Sigma^{-1}d \in \mathbb{R}^p. \end{align*} Because $d \neq 0$ and $\Sigma$ is invertible, $b \neq 0$. Moreover $\Sigma b = d$, so for every $a \in \mathbb{R}^p_0$, \begin{align*} a^\top d = a^\top \Sigma b = (a,b)_\Sigma, \qquad a^\top \Sigma a = (a,a)_\Sigma. \end{align*} Thus \begin{align*} J(a) = \frac{(a,b)_\Sigma^2}{(a,a)_\Sigma}. \end{align*} [/step]

[step:Prove the Cauchy--Schwarz bound in the $\Sigma$ inner product] [claim:Cauchy--Schwarz inequality for $(\cdot,\cdot)_\Sigma$] For all $x,y \in \mathbb{R}^p$, \begin{align*} (x,y)_\Sigma^2 \leq (x,x)_\Sigma (y,y)_\Sigma. \end{align*} If $y \neq 0$, equality holds if and only if $x$ is a scalar multiple of $y$. [/claim] [proof] If $y = 0$, then both sides are $0$. Assume $y \neq 0$. For every $t \in \mathbb{R}$, positive definiteness gives \begin{align*} 0 \leq (x - ty, x - ty)_\Sigma = (x,x)_\Sigma - 2t(x,y)_\Sigma + t^2(y,y)_\Sigma. \end{align*} Since $(y,y)_\Sigma > 0$, substitute \begin{align*} t := \frac{(x,y)_\Sigma}{(y,y)_\Sigma}. \end{align*} Then \begin{align*} 0 \leq (x,x)_\Sigma - \frac{(x,y)_\Sigma^2}{(y,y)_\Sigma}. \end{align*} Multiplying by $(y,y)_\Sigma > 0$ gives \begin{align*} (x,y)_\Sigma^2 \leq (x,x)_\Sigma (y,y)_\Sigma. \end{align*} Equality holds exactly when \begin{align*} (x - ty, x - ty)_\Sigma = 0. \end{align*} By positive definiteness, this is equivalent to $x - ty = 0$, hence to $x$ being a scalar multiple of $y$. [/proof] Applying the claim with $x := a$ and $y := b$ gives \begin{align*} (a,b)_\Sigma^2 \leq (a,a)_\Sigma (b,b)_\Sigma. \end{align*} Since $a \in \mathbb{R}^p_0$, positive definiteness gives $(a,a)_\Sigma > 0$, so division by $(a,a)_\Sigma$ yields \begin{align*} J(a) = \frac{(a,b)_\Sigma^2}{(a,a)_\Sigma} \leq (b,b)_\Sigma. \end{align*} Finally, \begin{align*} (b,b)_\Sigma = b^\top \Sigma b = (\Sigma^{-1}d)^\top \Sigma(\Sigma^{-1}d) = d^\top \Sigma^{-1}d, \end{align*} where the last equality uses $\Sigma^\top = \Sigma$. [/step]

[step:Identify exactly when equality occurs] The previous step shows that every $a \in \mathbb{R}^p_0$ satisfies \begin{align*} J(a) \leq d^\top \Sigma^{-1}d. \end{align*} Equality holds exactly when equality holds in the Cauchy--Schwarz inequality for $x := a$ and $y := b$. Since $b = \Sigma^{-1}d \neq 0$, this is equivalent to the existence of $\lambda \in \mathbb{R}$ such that \begin{align*} a = \lambda b = \lambda \Sigma^{-1}d. \end{align*} The condition $a \in \mathbb{R}^p_0$ forces $\lambda \neq 0$. Conversely, if $a = \lambda \Sigma^{-1}d$ with $\lambda \in \mathbb{R}\setminus\{0\}$, then \begin{align*} J(a) = \frac{(\lambda b^\top d)^2}{\lambda^2 b^\top \Sigma b} = \frac{(b^\top d)^2}{b^\top \Sigma b}. \end{align*} Since $d = \Sigma b$, we have $b^\top d = b^\top \Sigma b$, and therefore \begin{align*} J(a) = b^\top \Sigma b = d^\top \Sigma^{-1}d. \end{align*} Thus the maximisers are precisely the non-zero scalar multiples of $\Sigma^{-1}d$, and the maximum value is $d^\top \Sigma^{-1}d$. [/step]