# Show that $\sum_{\alpha \in \mathcal{A}} |\langle e_\alpha, x\rangle \langle e_\alpha,y \rangle| \leq \|x\| \cdot \|y\|$

The question is the following

Let $$S= \{e_\alpha:\alpha \in \mathcal{A}\}$$ be an orthonormal set in an inner space $$X$$. Show that for any $$x,y \in X$$ $$\sum_{\alpha \in \mathcal{A}} |\langle e_\alpha, x\rangle \langle e_\alpha,y \rangle| \leq \|x\| \cdot \|y\|$$

My attempt:

I tried to follow from the proof of Bessel's inequality, and here is what I did:

Let $$S = \{e_\alpha:\alpha \in \mathcal{A}\}$$ be an orthonormal set in an inner product space $$X$$. For any $$x,y \in X$$, take a finite subset $$\mathcal{A}' \subset \mathcal{A}$$ and compute the following: \begin{align*} &\left\langle \left(x-\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha x\rangle e_\alpha\right),\ \left(y-\sum_{\beta \in \mathcal{A}'} \langle e_\alpha y\rangle e_\alpha\right) \right\rangle \\ =\ &\langle x,y\rangle-\sum_{\beta \in \mathcal{A}'} \langle e_\beta, y\rangle \langle x, e_\beta \rangle -\sum_{\alpha \in \mathcal{A}'} \overline{\langle e_\alpha x\rangle} \langle e_\alpha, y \rangle + \sum_{\alpha,\beta \in \mathcal{A}'}\overline{\langle e_\alpha x\rangle} \langle e_\beta, y \rangle \langle e_\alpha, e_\beta \rangle \\ =\ &\langle x,y\rangle-\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha, y\rangle \langle x, e_\alpha \rangle \end{align*} Therefore $$\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha, y\rangle \langle x, e_\alpha \rangle = \langle x,y\rangle-\left\langle \left(x-\sum_{\alpha \in \mathcal{A}'} \langle e_\alpha x\rangle e_\alpha\right),\ \left(y-\sum_{\beta \in \mathcal{A}'} \langle e_\alpha y\rangle e_\alpha\right) \right\rangle$$ Here I wanted to apply the Cauchy-Schwarz inequality: $$|\langle x,y\rangle| \leq \|x\|\cdot \|y\|$$ but I don't know how to apply the absolute value. Or is there any simpler way to do it? Any help is appreciated.

$$\sum_{\alpha \in \mathcal A} |\langle e_{\alpha} , x \rangle| |\langle e_{\alpha}, y \rangle| \leq \sqrt {\sum_{\alpha \in \mathcal A}|\langle e_{\alpha} , x \rangle|^{2}} \sqrt {\sum_{\alpha \in \mathcal A}|\langle e_{\alpha} , y \rangle|^{2}}$$ from which the result follows.