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?


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

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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.