Let $E$ be a complex Hilbert space and $(A_1,...,A_n) \in \mathcal{L}(E)^n$.
If $(\lambda_1,\cdots,\lambda_n)\in \mathbb{C}^n$, why $$\left\|\sum_{k=1}^n\lambda_kA_k\right\|\leq \left(\sum_{k=1}^n|\lambda_k|^2\right)^{1/2}\left\|\sum_{k=1}^nA_kA_k^*\right\|^{1/2}$$
Take $x\in E$. Then applying Cauchy-Schwarz and elementary calculations we get $$ \begin{split} \| \sum_k \lambda_k A_kx\|^2 &= \langle (\sum_j \lambda_jA_j)x,\ (\sum_k \lambda_k A_k)x\rangle\\ &= \sum_{j,k} \lambda_j \bar\lambda_k\langle A_jx,A_kx\rangle\\ &\le \sum_{j,k} |\lambda_j|\cdot |\lambda_k| \cdot \|A_jx\|\cdot \|A_kx\|\\ &=\left( \sum_j |\lambda_j| \|A_jx\| \right)^2\\ &\le\left( \sum_j |\lambda_j|^2 \right) \left( \sum_j \|A_jx\|^2 \right)\\ &\le \left( \sum_j |\lambda_j|^2 \right) \|x\|^2 \cdot \|\sum_j A_j^*A_j\|, \end{split} $$ which proves the claim.