Find an example which shows that the following inequality is sharp Let $E$ be a complex Hilbert space. 
In (arXiv) it was shown that for $A=(A_1,...,A_n) \in \mathcal{B}(E)^n$ we have,
$$\displaystyle\frac{1}{2\sqrt{n}}\|A\|\leq \omega(A) \leq \|A\|,$$
where
$$
\omega(A) = \sup_{\|x\|=1} \left(\sum_{k=1}^n |\langle A_kx,x\rangle|^2\right)^{1/2},
$$
and
$$\|A\|= \left\|\displaystyle\sum_{k=1}^nA_k^*A_k \right\|^{1/2}.$$

How can we prove that $\frac{1}{2\sqrt{n}}$ is optimal?

 A: The estimate $\omega_e(A) \leq \Vert A \Vert$ is sharp, as we can take $A=(Id, \dots, Id)$ and for this choice we get equality.
We now prove optimality under the additional assumption $\dim_\mathbb{C}(E)\geq n+1$. In case someone has some insight for lower dimension please let me know.
If we assume $\dim_\mathbb{C}(E)\geq n+1$, then it suffices to consider the case $E=\mathbb{C}^{n+1}$ with the standard scalar product (otherwise choose a subspace of dimension $n+1$ and identify it with $\mathbb{C}^{n+1}$). We choose $A_k: \mathbb{C}^{n+1} \rightarrow \mathbb{C}^{n+1}$ such that
$$ A_k (x_1, \dots, x_{n+1})= (0, \dots,0 , x_1, 0, \dots, 0) $$
where $x_1$ is in the kth slot. The corresponding matrix consists of all zeros and a one in the kth row of the first column, e.g. for $n=3, k=2$
$$ A_2 = \begin{pmatrix}
0 & 0 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}.$$
Now we set
$$ A=(A_2, \dots, A_{n+1} )$$
We have
$$ \Vert A_k x \Vert^2 = \langle (0, \dots, x_1, \dots, 0),(0, \dots, x_1, \dots, 0)\rangle = \vert x_1 \vert^2  $$
Thus, we get
$$ \Vert A \Vert = \sup_{\Vert x \Vert = 1} \sqrt{n} \vert x_1 \vert = \sqrt{n} $$
On the other hand we have
$$ \vert \langle A_k x, x \rangle \vert^2 = \vert \langle (0, \dots, x_1, \dots, 0), (x_1, \dots, x_{n+1}) \rangle \vert^2 = \vert x_1 \vert^2 \cdot \vert x_k \vert^2  $$
And hence, for $\Vert x \Vert=1$
$$ \left(\sum_{k=2}^{n+1} \vert \langle A_k x, x \rangle \vert^2\right)^\frac{1}{2}
= \left(\vert x_1 \vert^2 \cdot \sum_{k=2}^{n+1} \vert x_k \vert^2 \right)^\frac{1}{2}
= \left(\vert x_1 \vert^2 \cdot (1- \vert x_1 \vert^2)\right)^\frac{1}{2}
= \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2}$$
Hence, we get
$$ \omega_e(A) = \sup_{\Vert x \Vert=1} \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2} 
= \sup_{\vert x_1 \vert\leq 1} \vert x_1 \vert \cdot \sqrt{1- \vert x_1 \vert^2} 
= \frac{1}{2} $$
Thus, we finally get
$$ \frac{1}{2\sqrt{n}} \Vert A \Vert = \frac{1}{2\sqrt{n}} \cdot \sqrt{n} = \frac{1}{2} = \omega_e(A) $$
A: As stated, the example cannot be right. The three expressions in the inequality spit coefficients, so if $T/\sqrt n$ works, so does $T$. 
Also the left inequalities is not sharp. For instance, if $E=\mathbb C$, then 
$$
\omega(A_1,\ldots,A_n)=\left\|\sum_{j=1}^n A_j^*A_j\right\|^{1/2}.
$$
