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.