# Proof of ternary "Cauchy-Schwarz" inequality $\left(\sum a_i b_i c_i\right)^2\le\sum a_i ^2\sum b_i ^2\sum c_i ^2$?

Prove the following inequality. $$\left( \sum a_i b_i c_i \right)^2 \leq \left( \sum a_i ^2 \right ) \left( \sum b_i ^2 \right) \left( \sum c_i ^2 \right ). \$$

As, it seems to be similar to Cauchy-Schwarz inequality, I thought of trying by applying the Schwarz inequality twice, by doing some kind of multiplication with $\left( \sum c_i ^2 \right )$ on either side, but result still seems to be far off reach.

Note: There is no restriction to use the Cauchy-Schwarz inequality.

• If you apply CS once you get $\sum a_{i}^{2} \sum b_{i}^{2} c_{i}^{2}$. Now consider bounding $\sum b_{i}^{2} c_{i}^{2}$. You can bound that sum by $\max(b_i^2) \sum c_{i}^{2}$. Now $\max(b_i^2)\le \sum b_i^{2}$ and the conclusion should follow Oct 5, 2016 at 16:04

This is exercise 1.3 (b) in J. Michael Steele's The Cauchy-Schwarz Master Class. The trick is to introduce the variable $\hat c_k=c^2_k/(c_1^2+\cdots +c_n^2)$ and use the usual Cauchy-Schwarz inequality.