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.


$$ 1 = (\sum_{j=1}^{n} p_j)^2 \leq \left(\sum_{j=1}^{n} \sqrt{p_j a_j}\sqrt{p_j b_j}\right)^2 \leq \sum_{j=1}^{n}p_j a_j \sum_{j=1}^{n} p_j b_j $$

  • $\begingroup$ Hello and thanks. I don't understand how you get to $(\sum^{n}_{j=1}{\sqrt{p_j a_j} \sqrt{p_j b_j}})^2$, I mean it's the inequality but I can't really see it. Could you expand the process? $\endgroup$
    – bajotupie
    Mar 13 '19 at 0:10
  • $\begingroup$ In the first inequality I only use $p_j \leq p_j \sqrt{a_j b_j}$ which is true because $p_j \geq 0$ and $a_j b_j \geq 1$. The second one is Cauchy-Schwarz applied to vectors $u,v$ with entries $u_i = \sqrt{p_i a_i}$ and $v_i = \sqrt{p_i b_i}$. $\endgroup$
    – user159517
    Mar 13 '19 at 5:41

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