Let $E$ be a complex Hilbert space, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$ and let $\mathcal{B}(E)$ the algebra of all bounded linear operators from $E$ to $E$.


I want to show that for $(T\_1,...,T\_d) \in \mathcal{B}(E)^d$ we have:
$$\displaystyle\frac{1}{2}\left\|\displaystyle\sum\_{k=1}^dT\_k^\*T\_k \right\|^{1/2}\leq \omega(T\_1,\cdots,T\_d) \leq \left\|\displaystyle\sum\_{k=1}^dT\_k^\*T\_k \right\|^{1/2},$$
where
 $$\omega(T\_1,\cdots,T\_d)=\displaystyle\sup\_{\|x\|=1}\bigg(\displaystyle\sum\_{k=1}^d|\langle T\_kx\;|\;x\rangle|^2\bigg)^{1/2}.$$


And you for you help.



A numerical radius inequality
The inequality you have written cannot be proven.


Gelu Popescu has a big paper on multivariable generalizations of the numerical radius (Memoirs of the AMS, [arXiv](https://arxiv.org/pdf/math/0410492.pdf)).
He calls what you have above the Euclidian operator radius of a tuple:
$$
\omega\_e(T\_1,\dots, T\_d) = \sup\_{\|x\|=1} \left(\sum\_{k=1}^d |\langle T\_kx,x\rangle|^2\right)^{1/2}
$$
and he proves many things about this in section 2 of his paper. Moreover, Proposition 2.4 establishes that
$$
\frac{1}{2\sqrt n} \|[T\_1, \dots, T\_d]\| \leq \omega\_e(T\_1,\dots, T\_d) \leq \|[T\_1, \dots, T\_d]\|
$$
and the inequality is sharp, noting that $\|[T\_1, \dots, T\_d]\| = \|\sum\_{k=1}^d T\_k^\*T\_k\|^{1/2}$. The proof is developed throughout section 2 but is not too difficult to follow.


