Let $E$ be a complex Hilbert space.
Let $A=(A_1,...,A_n) \in \mathcal{L}(E)^n$, why we have $$w_e(A):=\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{i=1}^n|\langle A_ix\;,\;x\rangle|^2\bigg)^{1/2}= \displaystyle\sup_{\|x\|=1}\sup_{(\lambda_1,...,\lambda_n)\in B_n}\bigg|\sum_{i=1}^n \lambda_i\langle A_ix\;,\;x\rangle\bigg|\;?$$ with $B_n$ is the open unit ball of $\mathbb{C}^n$.
And you for you help.
In fact, for arbitrary complex numbers $\nu_1,\dots,\nu_n$, you have $$ \sup_{(\lambda_1,\dots,\lambda_n) \in B_n} \left|\sum_{j=1}^n \lambda_j \nu_j\right| = \sqrt{\sum_{j=1}^n |\nu_j|^2} =:\alpha. $$ To see this, use the Cauchy Schwarz inequality to get "$\leq$". For the reverse inequality, choose $\lambda_j = \overline{\nu_j}/(\alpha+\epsilon)$ for arbitrary $\epsilon >0$
In more fancy language, the above is the characterization of the $\ell^2$ norm by duality.
If you apply this with $\nu_j = \langle A_j x,x \rangle$, you get your claim. Thus, the claim is not really related to the operators, it is just a statement about the $\ell^2$ norm.