I understand common proofs for Cauchy-Schwarz, but not sure about the first step in this one, which is proof 4 from here
Let $A = \sqrt{a_1^2 + a_2^2 + \dots + a_n^2}$ and $B = \sqrt{b_1^2 + b_2^2 + \dots + b_n^2}$. By the arithmetic-geometric means inequality (AGI), we have
$$ \sum_{i=1}^n \frac{a_ib_i}{AB} \leq \sum_{i=1}^n \frac{1}{2} \left( \frac{a_i^2}{A^2} + \frac{b_i^2}{B^2} \right) = 1 $$
so that
$$ \sum_{i=1}^na_ib_i \leq AB \leq \sqrt{\sum_{i=1}^na_i^2} \sqrt{\sum_{i=1}^n b_i^2} $$
Which is Cauchy-Schwarz. Now this is all quite elegant, but how is the first equation RHS equal to 1? And how is it AGI? Shouldn't it be in this form:
$$ \sqrt[n]{\prod_{i=1}^n x_i} \leq \frac{1}{n} \sum_{i=1}^n x_i $$
There's probably something simple I'm missing...