Cauchy Inequality proof by Induction

The Cauchy inequality

(a1b1 + a2b2 + …+anbn)^2 ≤ (a1^2 + a2^2 + … + an^2) (b1^2 + b2^2 + … + bn^2)

is one of the most widely used and most fundamental inequalities in all of mathematics especially for Real and Complex Analysis and there is no doubt that anyone who have taken a course will definitely vouch for the same. So a typical plan for us to give you an insight to this beautiful relationship will be through standard examples and challenge problems in this series of articles. So the next thing that strikes us all is the origin of challenge problems which are drawn from one of the world's famous mathematical competitions, but more often a problem is chosen because it illustrates a mathematical technique of wide applicability.

So how does one go about proving Cauchy inequality in a fresh way? Obviously there cannot be any universal method, but there are some hints that almost always help. One of the best of these is to try to solve the problem by means of a specific principle or specific technique.

Here, for example, one might insist on proving Cauchy inequality just by algebra - or just by geometry, by trigonometry, or by calculus. Miraculously enough, Cauchy inequality is wonderfully provable, and each of these approaches can be brought to a successful conclusion.

If one takes a dispassionate look at Cauchy inequality, there is another principle that suggests itself. Any time one faces a valid proposition that depends on an integer n, there is a reasonable chance that mathematical induction will lead to a proof. Since none of the standard texts in algebra or analysis gives such a proof of Cauchy inequality, this principle also has the benefit of offering us a path to an original proof - provided, of course, that we find any proof at all. When we look at Cauchy inequality for n = 1, we see that the inequality is trivially true.

This is all we need to start our induction, but it does not offer us any insight. If we hope to find a serious idea, we need to consider n = 2 and, in this second case, Cauchy inequality just says

(a1b1 + a2b2)^2 ≤ (a1^2 + a2^2) (b1^2 + b2^2)

This is a simple assertion, and you may see at a glance why it is true. Still, for the sake of argument, let us suppose that this inequality is not so obvious. How then might one search systematically for a proof?

Plainly, there is nothing more systematic than expanding both sides to find the equivalent inequality

(a1b1)^2 + (a2b2)^2 + 2(a1a2b1b2) ≤ (a1b1)^2 + (a1b2)^2 + (a2b1)^2 + (a2b2)^2

This on simplifying leads to an obvious result proving the Cauchy inequality for n=2, i.e

(a1b2 - a2b1)^2 ≥ 0

Now that we have proved a nontrivial case of Cauchy inequality, we are ready to look at the induction step. If we let H(n) stand for the hypothesis that Cauchy inequality is valid for n, we need to show that H(2) and H(n) imply H(n+1). With this plan in mind, we do not need long to think of first applying the hypothesis H(n) and then using H(2) to stitch together the two remaining pieces. Specifically, we have

(a1b1 + a2b2 +... + anbn + an+1bn+1) = (a1b1 + a2b2 +... + anbn) + an+1bn+1

≥ (a1^2 + a2^2 +... + an^2)^1/2 (b1^2 + b2^2 +... + bn^2)^1/2 + an+1bn+1

≥ (a1^2 + a2^2 +... + an^2 + an+1^2)^1/2 (b1^2 + b2^2 +... + bn^2 + bn+1^2)^1/2

where in the first inequality we used the induction hypothesis H(n), and in the second inequality we used H(2) in the form

pq + an+1bn+1 ≤ (p^2 + an+1^2)^1/2 (q^2 + bn+1^2)^1/2

with the new variables

p = (a1^2 + a2^2 +... + an^2)^1/2 and q = (b1^2 + b2^2 +... + bn^2)^1/2

The only difficulty one might have finding this proof comes in the last step where we needed to see how to use H(2). In this case the difficulty was quite modest, yet it anticipates the nature of the challenge one finds in more sophisticated problems. The actual application of Cauchy inequality is never difficult, the challenge always comes from seeing where Cauchy inequality should be applied and what one gains from the application.

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