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

holds for any pair of lists of numbers, and , for . 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

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.

An expansion gives

.

By adding an equivalent expression to both sides, we see

.

This makes it easier to see that

.

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

.

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

,

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

.

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

.

So in summary,

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

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[…] a fourth proof of the Cauchy-Schwarz inequality. I have previously presented a first, second and third proof. The fourth proof involves some algebraic pyrotechnics where we measure the difference […]

What about the case i=1?

It’s been so long ago since I wrote this, but probably I left the case i=1 as self-evident.

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?

Typo. Thanks!