$\newcommand\norm[1]{\lVert#1\rVert}$Let $B\in \mathbb R^{n\times n}$ be a symmetric and positive definite matrix. Assume that $x\in \mathbb R^n$ is the solution of $Bx=w$ for some given $w\in \mathbb R^n$. If we approximate $x$ using the conjugate gradient method with the initial guess $x_0\in \mathbb R^n$, it is known that the $k^\text{th}$ iterate satisfies
$$\norm{x - x_k}_B\leq\left(\frac{\sqrt{\kappa}-1}{\sqrt{\kappa}+1}\right)^k\norm{x - x_0}_B$$
for $k=1,2,\dotsc$, where $\norm z_B^2=z^TBz$ and $\kappa=\mu_\text{max}/\mu_\text{min}$ is the condition number of $B$, i.e., it is the ratio of the largest and smallest eigenvalues of $B$. I want to
Show that $\mu_\text{min}^{\frac{1}{2}}\norm z\leq\norm z_B\leq\mu_\text{max}^{\frac{1}{2}}\norm z$ for all $z\in \mathbb R^n$, where $\norm\cdot$ is the standard Euclidean norm.
Derive an error estimate for $\norm{x-x_k}$ in terms of $\norm{x-x_0}$, the condition number $\kappa$ and the iteration index $k$.
I feel like the error estimate can be derived if part 1) is resolved, which is why I am interested in 1) in the first place. But I'm having difficulty going from $\norm{x-x_k}_B\leq\left(\frac{\sqrt{\kappa}-1}{\sqrt{\kappa}+1}\right)^k\norm{x-x_0}_B$ to 1)
