Consider a positive definite matrix $A \in R^{d\times d}$ with it's largest eigenvalue being less that $\frac{1}{\lambda}$. Also consider $k-1$ vectors in $R^{d}$, $x_1, \cdots, x_{k-1}$. I am reading a paper which essentially does the following steps:
\begin{equation} \begin{split} \|A\sum_{i=1}^{k-1}x_i\| &\leq \|A^{1/2}\|\|\sum_{i=1}^{k-1}x_i\|_{A}\\ & \leq \frac{1}{\sqrt{\lambda}}\sqrt{k}\bigl(\sum_{i=1}^{k-1}\|x_i\|^2_{A}\bigr)^{1/2} \end{split} \end{equation}
I could follow when they used $\|A^{1/2}\| \leq \frac{1}{\sqrt{\lambda}}$. But not sure how they got the first inequality and also how they used Cauchy-Schwarz inequality in the second inequality -- for CS inequality, I am mostly confused about how they dealt with norm with respect to $A$ for $\sum_{i=1}^{k-1}1 = k-1 < k$. Can anyone help me understand the steps?
For any vector $v$, $$\|Av\| = \|A^{1/2} A^{1/2} v\| \overset{(i)}{\le} \|A^{1/2}\| \|A^{1/2} v\| \overset{(ii)}{=} \|A^{1/2}\| \|v\|_A.$$ Inequality (i) is due to the definition of the operator norm $\|A^{1/2}\|$, and equality (ii) is due to the definition of the norm $\|\cdot\|_A$ (i.e. $\|v\|_A^2 = v^\top A v$).
$$\|\sum_{i=1}^{k-1} x_i\|_A^2 = \sum_{i=1}^{k-1} \sum_{j=1}^{k-1} x_i^\top A x_j \le \sum_{i=1}^{k-1}\sum_{j=1}^{k-1} \|x_i\|_A \|x_j\|_A = \left(\sum_{i=1}^{k-1} \|x_i\|_A\right)^2 \le (k-1) \sum_{i=1}^{k-1} \|x_i\|_A^2.$$