Prove $(\sum\limits_k a_k b_k)^2 \leq \sum\limits_k b_k a_k^2 \sum\limits_k b_k$ using Cauchy Schwarz Let $a,b$ be two $n$ dimensional vectors, we want to show that
$(\sum\limits_k a_k b_k)^2 \leq \sum\limits_k b_k a_k^2 \sum\limits_k b_k$
Recall the Cauchy Schwarz inequality is given as 
$|\langle x,y \rangle| \leq \|x\|\|y\|$
then let $x = a,y = b$
Then
$|\langle a,b \rangle| \leq \|a\|\|b\|$
= $|a^Tb| \leq \sqrt{a^Ta} \sqrt{b^Tb}$ 
= $|a^Tb|^2 \leq a^Ta b^Tb$ 
$\Rightarrow$ $|\sum\limits_k a_k b_k|^2 \leq \sum\limits_k a_k^2 \sum\limits_k b_k^2$
At this point how can we eliminate the absolute value sign for the first term and  massage the right hand side to what we wish to show?

$|x|^2 = (x)^2$. Also, 
$$ \left(\sum_k a_k b_k\right)^2 = \left|⟨ \sqrt b· a, \sqrt{b}\rangle \right|^2\overset{C-S}{\leq} \sum_k b_k a_k^2 \sum_k b_k  $$
(Note that we need $b\geq 0$.)
