Cauchy-Schwarz Inequality and series proof 
Let {$a_n$} and {$b_n$} be sequences such that $\sum_{n=1}^\infty a_n^2$ and $\sum_{n=1}^\infty b_n^2$ are convergent
Prove that $\sum_{n=1}^\infty a_nb_n \le (\sum_{n=1}^\infty a_n^2)^{\frac{1}{2}} (\sum_{n=1}^\infty b_n^2)^{\frac{1}{2}}$

I can see I have to use the Cauchy-Schwarz inequality. But not really sure how to start. Also I know $\sum_{n=1}^\infty a_nb_n$ converges absolutely from a previous part of the question.
 A: Hint: Let $a=(a_1,\dots,a_n)$ and $b=(b_1,\dots,b_n)$. What does Cauchy-Schwarz say? The dot product of $a\cdot b \leq (a\cdot a)^{1/2}(b\cdot b)^{1/2}$. Note, that the terms in the brackets correspond to your expressions. Then let $n\to \infty$.
A: $(a_n)$ and $(b_n)$ are given to be real sequences. Now for any $x \in \mathbb{R}$, since $a_1,b_1 \in \mathbb{R}$, we can conclude that $(a_1x+b_1)^2\ge 0$.
Similarly, $(a_ix+b_i)^2\ge 0\  \forall \ i \in \mathbb{N}$.
Consider the polynomial $p(x) = \sum_{i=1}^{\infty} (a_ix+b_i)^2$.
By the previous calculations, it is clear that $\forall x\in \mathbb{R},\  p(x)\ \ge 0$ and it could have at most one real root that too when  $(a_ix+b_i)^2= 0\  \forall \ i \in \mathbb{N}$.
We know if a real polynomial has atleast 1 (and hence eventually 2) complex root, then the discriminant must be non-positive. 
Therefore discriminant must be less than or equal to $0$. Hence,
$(\sum_{i=1}^{\infty} a_ib_i)^2 \\ \le 
(\sum_{i=1}^{\infty} a_i)^2(\sum_{i=1}^{\infty} b_i)^2 $
i.e. $(\sum_{i=1}^{\infty} a_ib_i) \\ \le 
((\sum_{i=1}^{\infty} a_i)^2)((\sum_{i=1}^{\infty} b_i)^2 )^{1/2}$
i.e.  $(\sum_{i=1}^{\infty} a_ib_i) \\ \le 
((\sum_{i=1}^{\infty} a_i)^2)^{1/2}((\sum_{i=1}^{\infty} b_i)^2 )^{1/2}$
