A Fun Christmas Cauchy Exploration

Happy Holidays! 🎄

I wanted to review some fun results in math to practice, so I thought I'd turn it into a fun Christmas blog post. I'm going to talk a little bit about the Cauchy-Schwarz Inequality. It's an exciting result, so stick around.

1.1 Cauchy-Schwarz Inequality: Statement

For any real numbers $(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$, the following holds:

$${\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2⩽\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)$$

with equality $⟺(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$ are proportional, i.e., $\mathrm{\exists }$ a constant $\gamma$ such that $(a_1,a_2,...,a_n)=(\gamma b_1,\gamma b_2,...,\gamma b_n)$.

1.2 Jolly Cauchy-Schwarz Inequality Proofs

I'll begin with my favorite (and I think most simple) way to prove the inequality, which leverages properties of quadratics and the discriminant.

Proof 1:

We can consider a quadratic polynomial $f$ in $x$:

$$f(x)=\left(\sum _{i=1}^n{a}_i^2\right)x^2-2\left(\sum _{i=1}^n{a}_i{b}_i\right)x+\left(\sum _{i=1}^n{b}_i^2\right)=\sum _{i=1}^n(a_i^2{x}^2-2x{a}_i{b}_i+b_i^2)$$

$$=\sum _{i=1}^n(a_{i}x-b_i{)}^2$$

Notice now that $f(x)⩾0$ for all $x\in \mathbb{R}⟹$the descriminant is negative, i.e.,

$${(-2\left[\sum _{i=1}^n{a}_i{b}_i\right])}^2-4\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)={\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2-\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)⩽0$$

$$⟹\boxed{{\displaystyle {\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2⩽\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)}}$$

Another straightforward proof builds upon the AM-GM equality:

Proof 2:

Let $A=\sqrt{a_1^2+a_2^2+...+a_n^2},B=\sqrt{b_1^2+b_2^2+...+b_n^2}$. Then, by applying the AM-GM inequality to the terms $\frac{a_i^2}{A^2}$ and $\frac{b_i^2}{B^2}$,

$$\sum _{i=1}^n\frac{a_i{b}_i}{AB}⩽\sum _{i=1}^n\frac{1}{2}(\frac{a_i^2}{A^2}+\frac{b_i^2}{B^2})=1$$

which gives

$$\sum _{i=1}^n{a}_i{b}_i⩽AB=\sqrt{a_1^2+a_2^2+...+a_n^2}\sqrt{b_1^2+b_2^2+...+b_n^2}$$

Then, squaring both sides,

$$\boxed{{\displaystyle {\left(\sum _{i=1}^n{a}_i{b}_i\right)}^2⩽\left(\sum _{i=1}^n{a}_i^2\right)\left(\sum _{i=1}^n{b}_i^2\right)}}$$

Short, this post is, but I wanted to share something, at least!

-Elijah