alternative for Schwarz inequality I wish to show that
$$\sum_{n=1}^{\infty}|\langle e_n, x\rangle||\langle e_n,y \rangle| \leq \|x\|\|y\|$$
where $\{e_n\}$ is an orthonormal system. But my professor wants me to prove it without using Cauchy-Schwarz (or Hölder) inequality. His sugestion is to use the obvious inequality $(a-b)^2 \geq 0$ for the sequences $a_n = |\langle e_n, x\rangle|$ and $b_n = |\langle e_n, y\rangle| $. It follows that, in general,
$$ \sum_{n=1}^{\infty}\{(a_n - b_n)^2\} \geq 0 \implies \sum_{n=1}^{\infty}\{a_n^2 + b_n^2 - 2a_nb_n\}\geq 0$$
is this right? if so, by Bessel's inequality (i'm allowed to use it lol), $\sum a_n^2 = \sum|\langle e_n, x\rangle|^2 \leq \|x\|$ and $\sum b_n^2 =\sum |\langle e_n, y\rangle|^2 \leq \|y\|^2$. With this, I get to the inequality:
$$ 2\sum_{n=1}^{\infty}|\langle e_n, x\rangle||\langle e_n,y \rangle| \leq  \|x\|^2 + \|y\|^2$$
and... this is not quite what I want. Did i do something wrong or i'm forgetting something? how can I proceeed or restart it? any help will be very appreciated!
thanks in advance.
 A: Here is an idea I have using Cauchy product (I don't know if you are allowed to use it) :
Because the series are absolutely convergent then with Cauchy product :
$$\left(\sum_{n=1}^{\infty} a_n b_n \right)^2 = \sum_{n=1}^{\infty}\sum_{k=1}^n (a_k b_{k}) (a_{n-k}b_{n-k}) $$
Moreover with what you have done using Bessel's inequality:
$$ ||x||^2||y||^2 \geq \left(\sum_{n=1}^{\infty} a_n^2 \right)\left(\sum_{n=1}^{\infty} b_n^2 \right)= \sum_{n=1}^{\infty}\sum_{k=1}^n a_k^2b_{n-k}^2$$
and using your professor's suggestion, for all $n \geq k \geq 1$ we have:
$$(a_kb_{n-k})(a_{n-k} b_{k}) \leq \dfrac{1}{2}(a_k^2 b_{n-k}^2+a_{n-k}^2b_k^2)$$
Therefore :
$$
\left(\sum_{n=1}^{\infty} a_n b_n \right)^2 =
\sum_{n=1}^{\infty}\sum_{k=1}^n a_k b_{n-k} a_{n-k}b_{k} \leq \sum_{n=1}^{\infty}\sum_{k=1}^n \dfrac{1}{2}(a_k^2 b_{n-k}^2+a_{n-k}^2b_k^2)
$$
Furthermore :
$$ \sum_{n=1}^{\infty}\sum_{k=1}^n \dfrac{1}{2}(a_k^2 b_{n-k}^2+a_{n-k}^2b_k^2) = \sum_{n=1}^{\infty} \dfrac{1}{2} \left(\sum_{k=1}^n a_k^2 b_{n-k}^2 + \sum_{k=1}^n a_{n-k}^2 b_{k}^2 \right) = \sum_{n=1}^{\infty}\sum_{k=1}^n a_k^2b_{n-k}^2$$
Hence:
$$ \left(\sum_{n=1}^{\infty} a_n b_n \right)^2\leq||x||^2||y||^2$$
Finally:
$$\sum_{n=1}^{\infty} |\langle e_n|x\rangle ||\langle e_n|y \rangle|\leq ||x|| ||y|| $$
