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Is it valid to extend Cauchy-Schwarz to the 3 variable case with the following proof:

$$ \sum_{i=1}^n(a_ib_i) c_i \leq \sqrt{(\sum_{i=1}^{n}a_i^2b_i^2)(\sum_{i=1}^{n}c_i^2) } \leq \sqrt{\sum_{i=1}^{n}a_i^2} \sqrt{\sum_{i=1}^{n}b_i^2} \sqrt{\sum_{i=1}^{n}c_i^2} $$

The first inequality simply applies Cauchy-Schwarz to the two variable case. The last inequality assumes $\sum_{i=1}^{n}a_i^2b_i^2\leq (\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^{n}b_i^2)$. Is this a valid assumption?

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Yes, and your proof works as you suggest. You can complete it by expanding the product $(\sum_{i=1}^n a_i^2)(\sum_{i=1}^n b_i^2)$ and observing that the terms are all nonnegative and include all the terms of $\sum_{i=1}^n a_i^2b_i^2$.

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  • $\begingroup$ ooooooh right!! thanks!! $\endgroup$
    – vega
    Sep 22, 2018 at 16:49
  • $\begingroup$ is there a name for this inequality? $\endgroup$
    – vega
    Sep 22, 2018 at 16:49
  • $\begingroup$ Not that I know of $\endgroup$
    – Brent Kerby
    Sep 22, 2018 at 17:00
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    $\begingroup$ This inequality can be found in Michael Steele - The Cauchy-Schwarz Master Class, Exercise 1.3 page 13. $\endgroup$
    – Felix Huber
    Mar 1, 2022 at 17:14

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