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 $$\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{i=1}^n|\langle T_ix,x\rangle|^2\bigg)\geq\displaystyle\sup_{\|x\|=1}\bigg(\displaystyle\sum_{i=1}^n\|T_ix\|^2\bigg)\;.$$
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 $A-\varepsilon$.
Remark: Note that this is an answer to the original version of the question, for general (normal) $T_j$'s; the OP has now added the assumption that the $T_j$ commute.
This is false, unless $d=1$, when it has an easy proof using the spectral theorem. For a counterexample, consider $$ T_1 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \quad T_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} . $$ The RHS of your inequality equals $\sqrt{2}$ (take $x=e_1$); on the LHS, $|\langle x, T_j x\rangle |\le \|T_j\|= 1$ for each summand individually for any $x$, so the only chance would be to make them both $=1$, but that's clearly not working since $\langle e_1, T_2 e_1\rangle = 0$.
