How to solve $\sup_{v \in \mathbb{R}^n, a^2 I \preceq A }\| v^T(vv^T+A)^{-1} \|$ where $\| \cdot \|$ is the operator norm How to show find a bound on
\begin{align}
\sup_{v \in \mathbb{R}^n,  a^2 I \preceq A }\| v^T(vv^T+A)^{-1} \|
\end{align}
where $A$ is symmetric $n\times n$ matrix?  The norm is the operator norm.
Things that I did:
Solution for $n=1$:
\begin{align}
\sup_{v \in \mathbb{R},  a^2  \le A } \frac{|v|}{v^2+A}= \sup_{  a^2  \le A } \frac{\sqrt{A}}{A+A}=\sup_{  a^2  \le A }\frac{1}{2 \sqrt{A}}= \frac{1}{2 a}
\end{align}
Lower bound for any $n$. By choosing $v=[v_1,0, \ldots, 0]$ and $A=a^2I$, then
\begin{align}
\sup_{v_1 \in \mathbb{R}^n,  a^2 I \preceq A }\| v^T(vv^T+A)^{-1} \| \ge  \sup_{v_1 \in \mathbb{R},  a^2 \le A } \frac{|v_1| }{v_1^2+a^2}= \frac{1}{2 a}. 
\end{align}
Edit: I wanted to clarify that any norm will do for an answer. No need to focus on the operator norm.
 A: Note that if $\|\cdot\|$ denotes the operator norm associated with the usual Euclidean norm, then we have $\|x^T\| = \|x\|$ for any column-vector $x$. Thus, the supremum can be rewritten as
$$
\sup_{v \in \Bbb R^n, A \succeq a^2 I} \|(vv^T + A)^{-1}v\|.
$$
With the Sherman-Morrison formula, we have
$$
(vv^T + A)^{-1} = A^{-1} - \frac{[A^{-1}v][A^{-1}v]^T}{1 + v^TA^{-1}v}.
$$
So, we have
$$
(vv^T + A)^{-1}v = A^{-1}v - \frac{v^TA^{-1}v}{1 + v^TA^{-1}v}A^{-1}v = \frac{1}{1 + v^TA^{-1}v}A^{-1}v.
$$
Thus, the optimization becomes
$$
\sup_{v \in \Bbb R^n, A \succeq a^2 I} \frac{\|A^{-1}v\|}{1 + v^TA^{-1}v}.
$$
Now, if we make the substitution $w = A^{-1}v$, we end up with the rewritten maximization
$$
\sup_{w \in \Bbb R^n, A \succeq a^2 I} \frac{\|w\|}{1 + w^TAw}.
$$
From there, we see that $w^TAw \geq a^2\|w\|^2$ leads to the upper bound
$$
\sup_{w \in \Bbb R^n, A \succeq a^2 I} \frac{\|w\|}{1 + w^TAw} \leq \sup_{w \in \Bbb R^n} \frac{\|w\|}{1 + a^2 \|w\|} = \frac 12 a^{-1}.
$$
This upper bound is attained by taking $A = a^2 I$ and any $w$ with $\|w\| = a^{-1}$.
