Let $(H,\langle\cdot,\cdot\rangle)$ be a Hilbert space and $A\colon H\to H$ a bounded linear Operator such that $\langle A\cdot,\cdot\rangle$ is an inner product. Let $(\Vert x\Vert_A:=\sqrt{\langle Ax,x\rangle},\ x\in H)$, and let $a\in H$.
If we have $x\in H$ such that $\langle Ax, a\rangle=\Vert a\Vert_A$ and $\Vert x\Vert_A=1$, can we find an expression for $\Vert x\Vert$ (where $\Vert\cdot\Vert$ is the norm induced by the former inner product) using only $\Vert\cdot\Vert$?
You can apply CSB to the inner product $\langle \cdot, \cdot\rangle_A =\langle A\cdot, \cdot\rangle$. We have $$\left |\langle Ax, a\rangle\right| = \left|\langle x, a\rangle_A\right| \stackrel{CSB}{\le} \underbrace{\|x\|_A}_{= 1} \|a\|_A = \|a\|_A$$
Hence equality holds in CSB so there exists $t \in \mathbb{C}$ such that $x = t a$.
Assuming $a \ne 0$ we have $$1 = \|x\|_A = \|ta\|_A = |t|\|a\|_A \implies |t| = \frac{1}{\|a\|_A}$$
We conclude
$$\|x\| = \|ta\| = |t|\|a\| = \frac{\|a\|}{\|a\|_A}$$
If we assume $A \ge 0$ then $A^{1/2}$ exists and is also positive so
$$\|a\|_A^2 = \langle Aa, a\rangle = \langle A^{1/2}A^{1/2}a, a\rangle = \langle A^{1/2}a, A^{1/2}a\rangle = \|A^{1/2}a\|^2$$
Thus we can write
$$\|x\| = \frac{\|a\|}{\|A^{1/2}a\|}$$