Prove the The Cauchy--Schwarz--Buniakowski inequality using the fact that $E[X]^2 \le E[X^2]$.

Question

So like the title says I want to prove this whole thing. I'm going to paste the whole question here:

So this is what I have so far:

\begin{align*} & \sum_{i=1}^{n} \frac{u_{i}^2}{k} * \frac{v_{i}^2}{u_{i}^2} > \bigg[\sum_{i=1}^{n}\frac{u_{i}^2}{k} * \frac{v_{i}}{u_{i}}\bigg]^2 \\\\ & \Longleftrightarrow \frac{1}{k} \sum_{i=1}^{n} u_{i}^2 * \frac{v_{i}^2}{u_{i}^2} > \bigg[\frac{1}{k}\sum_{i=1}^{n} u_{i}^2 * \frac{v_{i}}{u_{i}}\bigg]^2 \\\\ & \Longleftrightarrow \frac{1}{k} \sum_{i=1}^{n} u_{i}^2 * \frac{v_{i}^2}{u_{i}^2} > \frac{1}{k^2}\bigg[\sum_{i=1}^{n} u_{i}^2 * \frac{v_{i}}{u_{i}}\bigg]^2 \\\\ & \Longleftrightarrow \frac{1}{k} \sum_{i=1}^{n} {v_{i}^2} > \frac{1}{k^2}\bigg[\sum_{i=1}^{n} u_{i}^2 * \frac{v_{i}}{u_{i}}\bigg]^2 \\\\ \end{align*}

I have no idea where to go from here. Should I stop here and try to plug in some value for $K$? I am super lost and don't even know what an appropriate value of $K$ would be.

Edit: Messed up some Latex lol

Answer 1

Consider the random variable $X$ whose values are $\frac{v_i}{u_i}$ ($j\in\{1,\dots,n\}$) with probability $p_i=P(X=\frac{v_i}{u_i})=\frac{u_i^2}{K}$. For this to be a probability distribution, we need $p_i\ge 0$ and $\sum_{i=1}^n p_i=1$. This means that $K> 0$ and $K=\sum_{i=1}^n u_i^2$.

We have $E(X)=\sum_{i=1}^n \frac{v_i}{u_i}\frac{u_i^2}{K}=\frac{1}{K}\sum_{i=1}^n {u_iv_i}$ and $E(X^2)=\sum_{i=1}^n \frac{v_i^2}{u_i^2}\frac{u_i^2}{K}=\frac{1}{K}\sum_{i=1}^n v_i^2$, therefore

$$ E(X^2)\ge E(X)^2 \iff \frac{1}{K}\sum_{i=1}^n v_i^2 \ge \frac{1}{K^2}\left(\sum_{i=1}^n {u_iv_i}\right)^2 \iff K\sum_{i=1}^n v_i^2 \ge \left(\sum_{i=1}^n {u_iv_i}\right)^2 $$

that is $\left(\sum_{i=1}^n u_i^2\right)\left(\sum_{i=1}^n v_i^2\right) \ge \left(\sum_{i=1}^n {u_iv_i}\right)^2$