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So I have to prove this using the Cauchy-Shwarz Inequality. I'm going to paste the whole question here: enter image description 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.

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2 Answers 2

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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.

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    $\begingroup$ Thank you so much this helped a lot! $\endgroup$
    – Aidan
    Commented Mar 1, 2023 at 22:50
  • $\begingroup$ Glad it helped :-)) $\endgroup$
    – Hamdiken
    Commented Mar 1, 2023 at 22:51
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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.

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  • $\begingroup$ Thank you so much this makes so much sense! $\endgroup$
    – Aidan
    Commented Mar 1, 2023 at 22:50

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