I've been struggling with this method of proving $E[X^2] > E[X]^2$.

Question

So I have to prove this using the Cauchy-Shwarz Inequality. 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} \times \frac{v_{i}^2}{u_{i}^2} > \bigg[\sum_{i=1}^{n}\frac{u_{i}^2}{k} \times \frac{v_{i}}{u_{i}}\bigg]^2 \\\\ & \Longleftrightarrow \frac{1}{k} \sum_{i=1}^{n} u_{i}^2 \times \frac{v_{i}^2}{u_{i}^2} > \bigg[\frac{1}{k}\sum_{i=1}^{n} u_{i}^2 \times \frac{v_{i}}{u_{i}}\bigg]^2 \\\\ & \Longleftrightarrow \frac{1}{k} \sum_{i=1}^{n} u_{i}^2 \times \frac{v_{i}^2}{u_{i}^2} > \frac{1}{k^2}\bigg[\sum_{i=1}^{n} u_{i}^2 \times \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 \times \frac{v_{i}}{u_{i}}\bigg]^2 \\\\ & \Longleftrightarrow k \sum_{i=1}^{n} {v_{i}^2} > \bigg[\sum_{i=1}^{n} u_{i} \times v_{i}\bigg]^2 \\\\ \end{align*}

I have no idea where to go from here. There is supposed to be only one value of K for which this inequality works,which is I'm sure to make all probabilities sum to 1. I'm super lost on what I'm supposed to do next. Thank you for any help.

Answer 1

You can choose $K=\|u\|^2$ (Given that $\|u\|=\sqrt{u_1^2+u_2^2+...+u_n^2}$) since the norm of a vector can be considered as a constant.

Answer 2

To fix $K$ you should fix $$ 1 = \sum_i P(X = \frac{v_i}{u_i}) = \frac{\sum_i u_i^2}{K} $$

so that $$ K = \sum_i u_i^2 $$

Putting this in your last equation, you get the Cauchy-Schwartz inequality.