Prove Cauchy-Schwarz with AM-GM for three variables

Question

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?

Answer 1

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

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

Answer 2

Late to the party, but here is a slightly different proof.

First, a useful lemma:

Lemma. (Monotonicity of $\ell_p$ norms) If $1\leq r\leq s \leq \infty$, then $\|{x}\|_s \leq \|{x}\|_r$ for every $x\in \mathbb{R}^n$.

Once you have this, apply Cauchy–Schwarz twice: $$\begin{align} \sum_{i=1}^n a_i b_i c_i &\leq \left(\sum_{i=1}^n a_i^2 b_i^2\right)^{1/2} \left( \sum_{i=1}^n c_i^2\right)^{1/2} \tag{CS}\\ &\leq \left(\sum_{i=1}^n a_i^4\right)^{1/4} \left(\sum_{i=1}^n b_i^4\right)^{1/4} \left( \sum_{i=1}^n c_i^2\right)^{1/2} \tag{CS}\\ &\leq \left(\sum_{i=1}^n a_i^2\right)^{1/2} \left(\sum_{i=1}^n b_i^2\right)^{1/2} \left( \sum_{i=1}^n c_i^2\right)^{1/2} \tag{Monotonicity $\ell_4/\ell_2$}\\ \end{align}$$

Note that this generalizes to more than 3 sequences: $$\begin{align} \sum_{i=1}^n a_{1,i} \dots a_{k,i} &\leq \prod_{\ell=1}^k \left(\sum_{i=1}^n a_{\ell,i}^2\right)^{1/2} \end{align}$$

Answer 3

Posting this answer because I am not satisfied with any of the current ones, for two reasons:

1. The proofs in the other answers look more complicated than what the above problem deserves, in my opinion. Let me give a simpler solution. Let $s=\sqrt{\sum_{i=1}^n c_i^2}$ we have $$s\sqrt{\sum_{i=1}^na_i^2} \sqrt{\sum_{i=1}^n b_i^2} \ge s\sum_{i=1}^n|a_ib_i| = \sum_{i=1}^n|a_ib_i|s \ge \sum_{i=1}^n|a_ib_ic_i|, $$ where the first inequality follows from Cauchy-Schwarz and the second follows from $s\ge |c_i|\ \forall i$. QED.

2. The original motivation of OP was to extend Cauchy-Schwarz to three variables, and thus it would be beneficial to point out that a potentially more interesting inequality could be the following for non-negative variables: $$\sum_{i=1}^na_ib_i c_i \leq \sqrt[3]{\sum_{i=1}^na_i^3} \sqrt[3]{\sum_{i=1}^n b_i^3}\sqrt[3]{\sum_{i=1}^n c_i^3}.$$ This inequality is known as Hölder's inequality.