Show that $\sum_{n=1}^\infty |\langle x,a_n\rangle|^2 \le \|x\|^2$ Let $H$ be a Hilbert space. Let $(a_1,a_2,...)$ be an orthonormal sequence in $H$.
Let $x \in H$. Show that $$\sum_{n=1}^\infty |\langle x,a_n\rangle|^2 \le \|x\|^2$$
I was thinking about move $\|x\|$ form RHS to LHS to make $\displaystyle\bigg\|\frac{x}{\|x\|}\bigg\|=1$,but I don't think this is helpful.
Can anyone give me a hint or idea of how to prove it since I don't know how to start it.
 A: If $\{\boldsymbol{\varphi}_1, \boldsymbol{\varphi}_2, ... \}$ be an orthonormal system in
$\mathcal{H}$, then the Bessel's inequality is:
$$\sum_{j=1}^{\infty} |\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle|^2 \leq \lVert
 \mathbf{x} \rVert^2 \quad \text{for every } \mathbf{x}\in \mathcal{H}.$$
Proof: 
\begin{align*}
0 &\leq \|\mathbf{x} -\sum_{i=1}^{n}\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle\boldsymbol{\varphi}_j \|^{2}\\ 
&= \Big\langle \mathbf{x} -\sum_{i=1}^{n}\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle\varphi_j, \mathbf{x} -\sum_{i=1}^{n}\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle\varphi_j \Big \rangle \\
&= \lVert \mathbf{x} \rVert^2 - \sum_{i=1}^{n} \langle \mathbf{x},\boldsymbol{\varphi}_j \rangle \langle \boldsymbol{\varphi}_j,\mathbf{x} \rangle - \sum_{i=1}^{n} \overline{\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle} \langle \boldsymbol{\varphi}_j,\mathbf{x} \rangle  +\sum_{i=1}^{n}  \langle \boldsymbol{\varphi}_j,\mathbf{x} \rangle \overline{\langle \mathbf{x},\boldsymbol{\varphi}_j \rangle}            \\
&= \lVert \mathbf{x} \rVert^2 - \sum_{i=1}^{n} | \langle \mathbf{x},\boldsymbol{\varphi}_j \rangle |^2,  \quad \forall n.
\end{align*}
