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