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.

5 responses to “The Cauchy–Schwarz Inequality III

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

  2. What about the case i=1?

  3. 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?

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