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I want to extend CS from two to three variables. Here's a Cauchy-Schwarz proof with two variables, which is proof 4 from here

Let $A = \sqrt{a_1^2 + a_2^2 + \dots + a_n^2}$ and $B = \sqrt{b_1^2 + b_2^2 + \dots + b_n^2}$. By the arithmetic-geometric means inequality (AGI), we have

$$ \sum_{i=1}^n \frac{a_ib_i}{AB} \leq \sum_{i=1}^n \frac{1}{2} \left( \frac{a_i^2}{A^2} + \frac{b_i^2}{B^2} \right) = 1 $$

so that

$$ \sum_{i=1}^na_ib_i \leq AB =\sqrt{\sum_{i=1}^na_i^2} \sqrt{\sum_{i=1}^n b_i^2} $$

How would I extend this method for three variables, i.e. to get the following? $$ \sum_{i=1}^na_ib_i c_i \leq \sqrt{\sum_{i=1}^na_i^2} \sqrt{\sum_{i=1}^n b_i^2} \sqrt{\sum_{i=1}^n c_i^2} $$

Somehow I don't think it's as trivial as the first method, i.e. simply defining $C$ the same way does not seem to work. Maybe there is a better approach?

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

First inequality follows using AM $\geq$ GM for two variables( $a_ib_i $'s as one variable, and $c_i$'s as another), and the second one follows as $\sum_{i=1}^{n}a_i^2b_i^2\leq (\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^{n}b_i^2).$

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  • $\begingroup$ What is the name of the second inequality you applied? $\endgroup$ – vega Sep 22 '18 at 15:57
  • $\begingroup$ @vega Don't know if it has a name. But it's very easy to prove as follows: $(\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^{n}b_i^2)=\sum_{i=1}^{n}a_i^2b_i^2+\sum_{i\neq j}a_i^2b_j^2 \geq \sum_{i=1}^{n}a_i^2b_i^2$ $\endgroup$ – Surajit Sep 22 '18 at 19:34
  • $\begingroup$ yeah I wrote exactly that down later and figured it out. Thanks a lot! $\endgroup$ – vega Sep 22 '18 at 20:10

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