I'm going to take a functional approach to inequality proofs: Rather
than discuss inequalities from axiomatic basics, I want to show you
the *heuristics* involved in proving inequalities.

Here are a couple of basic rules which I'll use constantly.

1. You can add a number to both (or all sides) of an inequality.

2. You can multiply an inequality by a nonzero number --- but if the number you multiply by is negative, the inequality is reversed.

* Example.* Prove that .

The looks like it came from . I know that even powers are always . I'll start with and see if I can get the desired inequality:

* Example.* If , then . And if and , then . Is it true that if and , then ?

The statement is false. For example, and , but .

This result shows that you have to be careful in the rules you use to work with inequalities. Some "rules" which look obvious aren't correct.

In fact, the false result in the example can be "fixed" by
placing additional assumptions on a, b, c, and d. To prove the
correct result, I'll have to use *very* basic facts about
inequalities involving real numbers.

Here are some *axioms* for the standard order relation on . Everything is defined in terms of a subset , the positive real numbers.

1. For every real number x, either , , or .

2. The sum of positive real numbers is a positive real number.

3. The product of positive real numbers is a positive real number.

The order relation is defined in terms of . Think about a statement like " ". Another way to say this is: You add a {\it positive} number (namely 4) to 3 to get 7.

* Definition.* If , means that , where .

As usual, means , means or , and means .

Here is a "fixed" version of the incorrect rule in the last
example. The proof illustrates a standard approach in inequality
proofs involving the basic axioms: *Convert inequality statements
to equations and work with the equations.*

* Lemma.* If , , and , then .

* Proof.* Suppose , , and . Write

Then

, , and are positive, because each is the product of positive numbers. Hence, is positive. The equation above therefore shows that .

* Lemma.* If , then .

* Proof.* Since , I know that , where . Hence,

But since , this means that .

Plainly, the same proof works if addition is replaced with subtraction.

You can often prove an inequality by transforming or substituting in a known inequality. Unfortunately, there are infinitely many inequalities you can start with! You have to rely on your knowledge of mathematics, as well as common sense: If you're trying to prove an inequality in calculus, for example, it's natural to think of all the "calculus inequalities" you know.

* Example.* (* Using a known
trig inequality*) Prove that for all ,
.

The " " form of the inequality and the presence of the in the middle remind me of , so I'll start with that and do some algebra:

* Example.* (* Using an
integral inequality*) From calculus, you know that if f and g are
integrable functions and on , then

Use this inequality to prove that

I have

Applying the integral inequality, I get

* Example.* You can often "see" that
an inequality is true by drawing a picture. For example, draw the
graph of for , where n is
an integer greater than 1.

Divide the interval up into n equal pieces, and build a rectangle on each piece, using the left-hand endpoints of each subinterval to get the heights. As you can see from the picture, the rectangles all lie above the curve, so the sum of the rectangle areas will be greater than the area under the curve.

The first rectangle has base 1 and height 1, so its area is 1. The second rectangle has base 1 and height , so its area is . Continuing in this way, the last rectangle has base 1 and height , so its area is .

The area under the curve is .

Therefore,

This inequality is correct (and by the way, you can use it to see that the harmonic series diverges). But the argument I gave is not a completely rigorous proof.

I'm assuming that the picture accurately represents the situation. To
*prove* that this is the case takes some work. For example,
I'd need to prove that each rectangle really *does* lie above
the curve. This would involve noting that shows that the graph is decreasing, then
using this to prove that the left-hand endpoints give the maximum
value of on each subinterval.

Pictures can help you *see* or *remember* that
something is true, and sometimes a picture or a *heuristic
argument* is useful in teaching --- to avoid obscuring the idea
with technicalities. But you should never confuse a picture with a
rigorous proof!

* Example.* (* Using the Mean
Value Theorem*) Prove that for all , .

Let . Take and apply the Mean Value Theorem to f on the interval . The Mean Value Theorem implies that there is a number c such that and

Now , and , so . Thus,

Therefore, , so .

In some cases, you can use a known inequality to prove other inequalities.

* The Triangle Inequality.* Let . Then

The name "Triangle Inequality" comes from the corresponding inequality when x and y are vectors. In that case, it says that the sum of the lengths of two sides of a triangle ( ) is greater than or equal to the length of the third side ( ).

* The Arithmetic-Geometric Mean Inequality.* Let
, and suppose that . Then

If x and y are two nonnegative numbers, their *
arithmetic mean* ("average") is and their * geometric mean* is
--- hence the name of this inequality.

* The Cauchy-Schwarz Inequality.* If , then

You may have seen this inequality in a vector calculus course or a linear algebra course. Let

Then the vector form of the inequality is

(The product on the left side is the dot product of the two vectors.)

* Example.* (* Using the
Triangle Inequality*) Prove that if a and b are real numbers,
then

Apply the Triangle Inequality with and :

Apply the Triangle Inequality with and :

The last step follows from the fact that .

Now

I've show that in both of these two cases, . Therefore, for all .

* Example.* (* Using the
Cauchy-Schwarz Inequality*) Prove that if , then

Apply the Cauchy-Schwarz Inequality with

I get

This simplifies to

But , so

As in this example, the trick to applying known inequalities is figuring out what substitutions to make.

Copyright 2009 by Bruce Ikenaga