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?


3 Answers 3


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

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, 2018 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, 2018 at 19:34
  • $\begingroup$ yeah I wrote exactly that down later and figured it out. Thanks a lot! $\endgroup$
    – vega
    Sep 22, 2018 at 20:10

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: $$\begin{align} \sum_{i=1}^n a_{1,i} \dots a_{k,i} &\leq \prod_{\ell=1}^k \left(\sum_{i=1}^n a_{\ell,i}^2\right)^{1/2} \end{align}$$

  • $\begingroup$ Isn't this a bit overkilled? Check my answer for a much simpler proof, which also works for the general case. $\endgroup$
    – f10w
    Sep 30, 2021 at 19:47
  • 1
    $\begingroup$ It is overkill, but lets the OP learn a new trick, maybe (monotonicity of norms); also, in the process establishes the (stronger) $\|abc\|_1 \leq \|a\|_4\|b\|_4\|c\|_2$, similar to both the $\|a\|_2\|b\|_2\|c\|_\infty$ and the $\|a\|_3\|b\|_3\|c\|_3$ one gets (implicitly or not) from your answer. One can check that any triple $r,s,t\leq 1$ with $1/r+1/s+1/t=1$ would do. @Khue $\endgroup$
    – Clement C.
    Sep 30, 2021 at 20:38
  • $\begingroup$ Yes, I just wanted to point out an elegant solution but I agree that all solutions are good solutions because of the information they contain (although my preference is always "simple solutions to simple problems"). +1 because of that. Cheers! $\endgroup$
    – f10w
    Oct 1, 2021 at 8:00

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.


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