# Demonstration using Cauchy-Schwarz inequality

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.

$$1 = (\sum_{j=1}^{n} p_j)^2 \leq \left(\sum_{j=1}^{n} \sqrt{p_j a_j}\sqrt{p_j b_j}\right)^2 \leq \sum_{j=1}^{n}p_j a_j \sum_{j=1}^{n} p_j b_j$$
• 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? Mar 13 '19 at 0:10
• 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 $v_i = \sqrt{p_i b_i}$. Mar 13 '19 at 5:41