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. 

Set for first $a_i = c_i^2$ and $b_i = d_i^2$, with $c_i,b_i\geq 0$. We know that:
$$\left(\sum_{i=1}^{n}c_i d_i\right)^2 \leq \left(\sum_{i=1}^{n}c_i^2\right)\left(\sum_{i=1}^{n}d_i^2\right)\tag{1}$$
and we need to prove that:
$$\left(\sum_{i=1}^{n+1}c_i d_i\right)^2 \leq \left(\sum_{i=1}^{n+1}c_i^2\right)\left(\sum_{i=1}^{n+1}d_i^2\right).\tag{2}$$
We have:
$$\begin{eqnarray*}\left(\sum_{i=1}^{n+1}c_i d_i\right)^2 &=& \left(\sum_{i=1}^{n}c_i d_i\right)^2+(c_{n+1}d_{n+1})^2\\ &+& 2c_{n+1}d_{n+1}\left(\sum_{i=1}^{n}c_i d_i\right),\end{eqnarray*}$$
$$\begin{eqnarray*}\left(\sum_{i=1}^{n+1}c_i^2\right)\left(\sum_{i=1}^{n+1}d_i^2\right)&=&\left(\sum_{i=1}^{n}c_i^2\right)\left(\sum_{i=1}^{n}d_i^2\right)+(c_{n+1}d_{n+1})^2\\&+&c_{n+1}^2\sum_{i=1}^n d_i^2 + d_{n+1}^2\sum_{i=1}^n c_i^2\end{eqnarray*}$$
hence in order to prove $(2)$ we just need to show that
$$2c_{n+1}d_{n+1}\left(\sum_{i=1}^{n}c_i d_i\right)\leq c_{n+1}^2\sum_{i=1}^n d_i^2 + d_{n+1}^2\sum_{i=1}^n c_i^2.\tag{3}$$
Since 
$$c_{n+1}^2\sum_{i=1}^n d_i^2 - 2c_{n+1}d_{n+1}\left(\sum_{i=1}^{n}c_i d_i\right)+ d_{n+1}^2\sum_{i=1}^n c_i^2$$
regarded as a bivariate polynomial in $c_{n+1}$ and $d_{n+1}$, has a non-positive discriminant due to $(1)$, $(3)$ is proved.
