Let $(T\_1,..., T\_n)\in \mathcal{L}(E)^n$ be a tuple of commuting normal operators (i.e. each $T\_k$ is normal and $T\_iT\_j=T\_jT\_i$ for all $i,j$), where $E$ is a complex Hilbert space.


I want to show that
$$\displaystyle\sup\_{\|x\|=1}\bigg(\displaystyle\sum\_{i=1}^n|\langle T\_ix,x\rangle|^2\bigg)\geq\displaystyle\sup\_{\|x\|=1}\bigg(\displaystyle\sum\_{i=1}^n\|T\_ix\|^2\bigg)\;.$$



Inequality for normal operators
Consider the case of finite dimensional Hilber space. Then may suppose that operators $T\_1,\dots,T\_d$ are diagonal in the same orthonormal basis $(e\_1,e\_2,\dots)$, denote the diagonal of $T\_i$ by $(p\_{i1},p\_{i2},\dots)$. The square of RHS is nothing but $\sup\_j |p\_{1j}|^2+|p\_{2j}|^2+\dots+|p\_{dj}|^2$. For fixed $j$, taking $x=e\_j$ we get the same thing $|p\_{1j}|^2+|p\_{2j}|^2+\dots+|p\_{dj}|^2$ for LHS.


Now do the same in general situation. Note that 
$$
\sup\_{\|f\|\_{L^2(\mu)}=1}\bigg[\displaystyle\sum\_{i=1}^n\left(\int\_Y|\varphi\_i|^2|f|^2d\mu \right)\bigg]={\rm essup}\,\sum\_i |\varphi\_i|^2=:A.
$$
Note that (by separability of a complex plane and by the definition of essential supremum) we may find discs $D\_i$ such that $\mu(\Omega)>0$, where $\Omega=\{x:\varphi\_i(x)\in D\_i,i=1,\dots,n\}>0$ and $\sum\_i |\theta\_i|^2>A-\varepsilon$ whenever $\theta\_i\in D\_i$. Choose any function $f$ with $\|f\|=1$ concentrated in $\Omega$. Then $\int \varphi\_i |f|^2d\mu\in D\_i$ (indeed, $D\_i$ are convex, thus a weighted mean of elements of $D\_i$ belongs to $D\_i$). It implies that LHS is not less than $A-\varepsilon$.


