The Cauchy–Schwarz Inequality III

I have previously written about the Cauchy-Schwarz Inequality here and here. To recap, the Cauchy-Schwarz inequality,

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

holds for any pair of lists of $n$ numbers, ${a_i}$ and ${b_i}$ , for $i=1,2,\ldots,n$ . Now, I have given two proofs so far, and here is a third. We will proceed by induction on the number of variables. A second form of the Cauchy-Schwarz inequality is

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

I have previously explained why this is another form of the Cauchy-Schwarz inequality.

Now for the base case of induction, we can show that the Cauchy-Schwarz holds for lists of 2 numbers. We start as always with the fact that squares of real numbers are always non-negative.

$(a_1b_2 - a_2b_1) ^2\geq 0.$

An expansion gives

$a_1^2b_2^2 +a_2^2b_1^2 \geq 2a_1^2b_1^2a_2^2b_2^2$ .

By adding an equivalent expression to both sides, we see

$a_1^2b_1^2 + a_1^2b_2^2 +a_2^2b_1^2 + a_2^2b_2^2\geq a_1^2b_1^2 + 2a_1^2b_1^2a_2^2b_2^2 + a_2^2b_2^2$ .

This makes it easier to see that

$\left(a_1^2+a_2^2\right)\left(b_1^2+b_2^2\right)\geq \left(a_1b_1+a_2b_2\right)^2$ .

Suppose the Cauchy-Schwarz inequality is true for some lists of numbers of length $k$ , that is,

$\displaystyle\sum_{i=1}^k{a^2_i}\sum_{i=1}^k{b^2_i} \geq \sum_{i=1}^k{a_ib_i}$ .

We can leverage this assumption for lists of length $k$ into a statement about lists of length $k+1$ . Starting with

$\displaystyle \left(\sum_{i=1}^{k+1}{a_ib_i}\right)^2 = \left(a_{k+1}b_{k+1}+\sum_{i=1}^k{a_ib_i}\right)^2$ ,

we can apply the assumption of the Cauchy-Schwarz inequality for lists of length $k$ to deduce that

$\displaystyle\left(a_{k+1}b_{k+1}+\sqrt{\sum_{i=1}^k{a_i^2}}\sqrt{\sum_{i=1}^k{b_i^2}}\right)^2 \geq \left(a_{k+1}b_{k+1}+\sum_{i=1}^k{a_ib_i}\right)^2$ .

Applying the Cauchy-Schwarz inequality for $n=2$ , we can deduce that the left side of the last expression is less that or equal to

$\displaystyle\left(a_{k+1}^2 + \left(\sqrt{\sum_{i=1}^k{a_i^2}}\right)^2\right)\left(b_{k+1}^2 + \left(\sqrt{\sum_{i=1}^k{b_i^2}}\right)^2\right)$ .

So in summary,

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

That is to say, as promised, if the Cauchy-Schwarz inequality is true for $k$ then it is true for $k+1$ . Therefore, the Cauchy-Schwarz inequality holds for lists of numbers of any length.

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5 Responses to The Cauchy–Schwarz Inequality III

Pingback: The Cauchy–Schwarz Inequality IV « The Twofold Gaze
Anonymous | November 21, 2012 at 11:34 am | Reply

What about the case i=1?
- Kareem Carr | December 24, 2012 at 9:17 pm | Reply
  
  It’s been so long ago since I wrote this, but probably I left the case i=1 as self-evident.
Jeroen | November 28, 2012 at 6:39 am | Reply

When you apply the assumption of the Cauchy-Schwarz inequality for lists of length k, why is the summation on the left side of the expression till n, and not till k?
- Kareem Carr | December 24, 2012 at 9:17 pm | Reply
  
  Typo. Thanks!

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The Cauchy–Schwarz Inequality III

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