Proving the Cauchy-Schwarz inequality by induction I ran across this problem in some old notes, and I frustratingly can't figure out how to do it 
Let $a_i$ and $b_i$ be sequences of natural numbers, use induction to show
$\sum_{i=1}^n (a_ib_i)^{1/2} \le (\sum_{i=1}^n a_i)^{1/2}(\sum_{i=1}^n b_i)^{1/2} $
Obviously this is trivial to show for n=1. I can't make much progress on n+1. I've tried various tactics, squaring both sides etc. 
Any hint or help would be appreciated. Thanks. 

Key Step:
$a_1b_1+a_2b_2+...+a_nb_n+a_{n+1}b_{n+1} $
$= (a_1b_1+a_2b_2+...+a_nb_n)+a_{n+1}b_{n+1}$
$ \le \left(a_1^2+a_2^2+...+a_n^2 \right)^{1/2}\left(b_1^2+b_2^2+...+b_n^2\right)^{1/2}+a_{n+1}b_{n+1}$
$\le \left( a_1^2+a_2^2+...+a_n^2+a_{n+1}^2 \right)^{1/2} \left( b_1^2+b_2^2+...+b_n^2+b_{n+1}^2 \right)^{1/2}$
where in the first inequality we used the induction hypothesis, and in the second
inequality we use the case n = 2 in the form
$\alpha \beta +a_{n+1}b_{n+1} \le (\alpha^2+a_{n+1}^2)^{1/2}(\beta^2+b_{n+1}^2)^{1/2}$ with the new variables
$\alpha =\left( a_1^2+a_2^2+...+a_n^2 \right)^{1/2}$ and
$\beta = \left(b_1^2+b_2^2+...+b_n^2\right)^{1/2}$
