Suppose $0\leq p_j\leq 1$ for $j=1,2,3...n$, so that $p_1+...+p_n = 1$. Let's $a_j,b_j \geq 1$ so that $a_j b_j \geq1$ for $j=1,2,3...n$. Demonstrate:
$1 \leq \sum^{n}_{j=1}p_ja_j \sum^{n}_{j=1}p_jb_j$
This is to be solved using the Cauchy-Schwarz inequality: $|(u,v)|² \leq (u,u)(v,v)$
This is what I got so far:
$\sum^{n}_{j=1}p_ja_j \sum^{n}_{j=1}p_jb_j = (\vec{p},\vec{a})(\vec{p},\vec{b})$
$1 \leq |(\vec{p},\vec{a})||(\vec{p},\vec{b})|$
$1 \leq |(\vec{p},\vec{a})|²|(\vec{p},\vec{b})|²$
I don't know how to apply the initial conditions.
$$ 1 = (\sum_{j=1}^{n} p_j)^2 \leq \left(\sum_{j=1}^{n} \sqrt{p_j a_j}\sqrt{p_j b_j}\right)^2 \leq \sum_{j=1}^{n}p_j a_j \sum_{j=1}^{n} p_j b_j $$