# The Cauchy–Schwarz Inequality

I recently read The Cauchy-Schwarz Master Class by Michael Steele which, in addition to teaching the reader about inequalities, has a lot of fun and varied things to say about the Cauchy-Schwarz inequality.

First things, first, the Cauchy-Schwarz inequality in it’s simplest form says the following:

$\displaystyle\sqrt{\sum_{i=1}^n{a_i^2}}\sqrt{\sum_{i=1}^n{b_i^2}} \geq \sum_{i=1}^n{a_ib_i}$

This holds for any pair of lists of $n$ numbers, ${a_i}$ and ${b_i}$, for $i=1,2,\ldots,n$.  We know that the square of any number is greater than or equal to zero. Thus, for two real numbers,

$(a-b)\geq 0$

and

$a^2 - 2ab + b^2 \geq 0$;

but with some algebra,

$\displaystyle\frac{1}{2}\left(a^2+b^2 \right) \geq a b.$

We can easily apply this knowledge to the lists $a_i$ and $b_i$ for $i=1,2,\ldots,n$.

$\displaystyle\frac{1}{2}\left( \sum_{i=1}^n{a_i^2}+\sum_{i=1}^n{b_i^2}\right) \geq\sum_{i=1}^n{a_ib_i}$

Here we apply a rather entertaining trick. Suppose, we had two lists of numbers which we defined using the original lists as follows:

$\displaystyle\hat{a}_i = \frac{a_i}{\sqrt{\sum_{i=1}^n{a_i^2}}}$

$\displaystyle\hat{b}_i = \frac{b_i}{\sqrt{\sum_{i=1}^n{b_i^2}}}$

This would change the situation in this way,

$\displaystyle 1=\frac{1}{2}\left(1 + 1\right)=\frac{1}{2}\left( \sum_{i=1}^n{\hat{a}_i^2}+\sum_{i=1}^n{\hat{b}_i^2}\right)\geq\sum_{i=1}^n{\hat{a}_i\hat{b}_i}$.

So, it follows that

$\displaystyle 1\geq\sum_{i=1}^n{\frac{a_i}{\sqrt{\sum_{i=1}^n{a_i^2}}}\frac{b_i}{\sqrt{\sum_{i=1}^n{b_i^2}}}}$.

In summary,

$\displaystyle\sqrt{\sum_{i=1}^n{a_i^2}}\sqrt{\sum_{i=1}^n{b_i^2}} \geq \sum_{i=1}^n{a_ib_i}$.

This concludes the proof of the Cauchy-Schwarz inequality.