Advances in Pure Mathematics
Vol.3 No.3(2013), Article ID:31228,4 pages DOI:10.4236/apm.2013.33047
An Elementary Proof of the Mean Inequalities
Department of Statistics, George Mason University, Fairfax, USA
Email: iizmirl2@gmu.edu
Copyright © 2013 Ilhan M. Izmirli. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Received November 24, 2012; revised December 30, 2012; accepted February 3, 2013
Keywords: Pythagorean Means; Arithmetic Mean; Geometric Mean; Harmonic Mean; Identric Mean; Logarithmic Mean
ABSTRACT
In this paper we will extend the well-known chain of inequalities involving the Pythagorean means, namely the harmonic, geometric, and arithmetic means to the more refined chain of inequalities by including the logarithmic and identric means using nothing more than basic calculus. Of course, these results are all well-known and several proofs of them and their generalizations have been given. See [1-6] for more information. Our goal here is to present a unified approach and give the proofs as corollaries of one basic theorem.
1. Pythagorean Means
For a sequence of numbers we will let
and
to denote the well known arithmetic, geometric, and harmonic means, also called the Pythagorean means.
The Pythagorean means have the obvious properties:
1) is independent of order 2)
3)
4) is always a solution of a simple equation. In particular, the arithmetic mean of two numbers and can be defined via the equation
The harmonic mean satisfies the same relation with reciprocals, that is, it is a solution of the equation
The geometric mean of two numbers and can be visualized as the solution of the equation
1)
2)
3)
This follows because
2. Logarithmic and Identric Means
The logarithmic mean of two non-negative numbers and is defined as follows:
and for positive distinct numbers and
The following are some basic properties of the logarithmic means:
1) Logarithmic mean can be thought of as the mean-value of the function over the interval.
2) The logarithmic mean can also be interpreted as the area under an exponential curve.
Since
We also have the identity
Using this representation it is easy to show that
1) We have the identity
which follows easily:
To define the logarithmic mean of positive numbers, we first recall the definition of divided differences for a function at points, denoted as
For
and for and,
We now define
So for example for n = 2, we get
The identric mean of two distinct positive real numbers is defined as:
with.
The slope of the secant line joining the points
and on the graph of the function
is the natural logarithm of.
It can be generalized to more variables according by the mean value theorem for divided differences.
3. The Main Theorem
Theorem 1. Suppose is a function with a strictly increasing derivative. Then
for all in.
Let be defined by the equation
Then,
is the sharpest form of the above inequality.
Proof. By the Mean Value Theorem, for all in, we have
for some between and. Assuming without loss of generality by the assumption of the theorem we have
Integrating both sides with respect to, we have
and the inequality of the theorem follows.
Let us now put
Note that
Moreover, since
there exists an in such that.
Since is strictly increasing, we have
for
and
for
Thus, is a minimum of and for all
4. Applications to Mean Inequalities
We will extend the well-known chain of inequalities
to the more refined
using nothing more than the mean value theorem of differential calculus. All of these are strict inequalities unless, of course, the numbers are the same, in which case all means are equal to the common value of the two numbers.
Let us now assume that
Let us let The condition of the Theorem 1 is satisfied. Solving the equation
we find
where
Hence the left-hand side of the inequality becomes
Thus we have
implying
or
Let us let. The condition of Theorem 1 is satisfied. We can easily compute the of the theorem from the equation
as
Our inequality becomes
Implying,
that is
Now let. Again the condition of Theorem 1 is satisfied. The of the theorem can be computed from the equation
as
where
Since
Thus,
where
Consequently our inequality becomes
implying
that is,
Finally, let us put. Again the condition of Theorem 1 is satisfied. Since in this case
the of the theorem can be computed as
The right-hand side of the inequality becomes
The integral on the left-hand side of our inequality yields
implying
or
Thus, we now have for
REFERENCES
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