Let $E$ be a complex Hilbert space.



> 
> Let $(A\_1,...,A\_n) \in \mathcal{L}(E)^n$, could you please help me to show that
>  $$\displaystyle\sup\_{\|x\|=1}\bigg(\displaystyle\sum\_{i=1}^n\| A\_i^\*x\|^2\bigg)\leq (4n)\displaystyle\sup\_{\|x\|=1}\bigg(\displaystyle\sum\_{i=1}^n|\langle A\_ix\;,\;x\rangle|^2\bigg).$$
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> 
> 


And you for you help.



inequality involving tuple of operators on Hilbert spaces
Notice that
$$ \|[A\_1 \dots A\_n]\| = \|[A\_1 \dots A\_n]^\*\| = \left\| \left[\begin{smallmatrix} A\_1^\* \\ \\\vdots \\ A\_n^\* \end{smallmatrix}\right] \right\| = \sup\_{\|x\|=1} \left( \sum\_{i=1}^n \|A\_i^\*x\|^2 \right)^{1/2}.$$


By the answer to [this question](https://mathoverflow.net/questions/285471/a-numerical-radius-inequality/285476#285476) we have that
$$ \frac{1}{2\sqrt n} \|[A\_1 \dots A\_n]\| \leq \sup\_{\|x\|=1} \left(\sum\_{i=1}^n |\langle A\_ix,x\rangle|^2\right)^{1/2}. $$


Combine these two statements to get your desired inequality.


