# Prove Cauchy-Schwarz with AM-GM for three variables

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 $$\sum_{i=1}^{n}a_i^2b_i^2\leq (\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^{n}b_i^2).$$
• @vega Don't know if it has a name. But it's very easy to prove as follows: $(\sum_{i=1}^{n}a_i^2)(\sum_{i=1}^{n}b_i^2)=\sum_{i=1}^{n}a_i^2b_i^2+\sum_{i\neq j}a_i^2b_j^2 \geq \sum_{i=1}^{n}a_i^2b_i^2$ – Surajit Sep 22 '18 at 19:34