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.

3 responses to “The Cauchy–Schwarz Inequality

  1. Pingback: The Cauchy–Schwarz Inequality II « The Twofold Gaze

  2. Pingback: The Cauchy–Schwarz Inequality III « The Twofold Gaze

  3. Pingback: The Cauchy–Schwarz Inequality IV « The Twofold Gaze

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