Let $X$ be an inner product space and $\{e_{n}\}_{n=1}^{\infty} \subset X$ be an orthonormal set. Show that $$ \sum_{n=1}^{\infty}|\langle x,e_{n}\rangle\langle y, e_{n}\rangle| \leqslant \|x\|\|y\| $$ for all $x,y \in X$.
I am seeking a hint on how to get started. I imagine this involves some manipulation of the $RHS$ to get down to inequality on the $LHS$. Any insight is appreciated.
Using the hints below:
$$ \sum_{n=1}^{\infty}|\langle x,e_{n}\rangle\langle y, e_{n}\rangle| \leqslant \sum_{n=1}^{\infty}\|x\|\|y\|$$ I do not quite see the connection to Bessel's inequality from here.
Cauchy Schwarz says that $$ \left(\sum_{n=1}^\infty\langle x,e_n\rangle\langle y,e_n\rangle\right)^2 \le\sum_{n=1}^\infty\langle x,e_n\rangle^2\sum_{n=1}^\infty\langle y,e_n\rangle^2\tag{1} $$ If we are in a Hilbert Space, we can apply Bessel's Inequality. If not, we might want to reprove it for an inner product space.
For any $1\le k\le n$, $$ \left\langle x-\sum_{j=1}^n\langle x,e_j\rangle e_j,e_k\right\rangle =0\tag{2} $$ If we set $$ u=\sum_{j=1}^n\langle x,e_j\rangle e_j\tag{3} $$ $(2)$ and $(3)$ show that $\langle u,x-u\rangle=0$. Therefore, $$ \begin{align} \|x\|^2 &=\langle x,x\rangle\\ &=\langle u+(x-u),u+(x-u)\rangle\\ &=\|u\|^2+\|x-u\|^2\\ &\ge\|u\|^2\\ &=\sum_{j=1}^n\langle x,e_j\rangle^2\tag{4} \end{align} $$ Letting $n\to\infty$, $(1)$ and $(4)$ show that $$ \left(\sum_{k=1}^\infty\langle x,e_k\rangle\langle y,e_k\rangle\right)^2 \le\|x\|^2\|y\|^2 $$