Cauchy-Schwartz inequality

Question

Let ${\bf T} = (T_1,...,T_d) \in \mathcal{B}(F)^d$, with $\mathcal{B}(F)$ denotes the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.

It is true that $$\left(\sum_{|\alpha|=n} \|{\bf T}^\alpha x\|\|{\bf T}^\alpha y\|\right)^2\leq \left(\sum_{|\alpha|=n} \|{\bf T}^\alpha x\|^2\right) \left(\sum_{|\alpha|=n} \|{\bf T}^\alpha y\|^2\right)\;?$$

with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{Z}_+^d,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$; and ${\bf T}^\alpha:=T_1^{\alpha_1} T_2^{\alpha_2}...T_d^{\alpha_d}$

Answer 1

Evidently we know C-S in this form:

CS1 If $x_j,y_j\ge0$ for $j=1,2\dots,n$ then $\left(\sum_1^n x_jy_j\right)^2 \le\sum_1^nx_j^2\sum_1^ny_j^2$

but we're stuck on this version:

CS2 If $F$ is a finite set and $x_\alpha,y_\alpha\ge0$ for $\alpha\in F$ then $\left(\sum_{\alpha\in F}x_\alpha y_\alpha\right)^2\le\sum_{\alpha\in F}x_\alpha^2\sum_{\alpha\in F}y_\alpha^2$.

It is awesomely obvious that the two are equivalent: To show that CS2 implies CS1, let $F=\{1,2,\dots,n\}$. To show that CS1 implies Cs2: Suppose that $F=\{\alpha_1,\dots,\alpha_n\}$. For any $X_\alpha$ defined for all $\alpha\in F$ we have $$\sum_{\alpha\in F}X_\alpha=\sum_{j=1}^n X_{\alpha_j}.$$(Apply that three times, with $X_\alpha=x_\alpha y_\alpha$, $X_\alpha=x_\alpha^2$ and $X_\alpha=y_\alpha^2$.)