![MathJax Logo](/templates/jsp/_style2/_tandf/pb2/images/math-jax.gif)
Abstract
We establish some inequalities comparing power means of two numbers with combinations of the arithmetic and geometric means. A conjecture from [Citation1] is confirmed.
MSC:
Given positive numbers a, b, the arithmetic, geometric and pth power means arefor
. With a, b fixed, we denote these just by A, G, and Mp.
Of course, these definitions extend in a natural way to more than two numbers. For any finite set of positive numbers, it is clear that , and well known that
for
,
for
and
for p < 0. (Also, one defines M0 to be G).
For two numbers, it is easily seen that . This equality does not extend to more than two numbers: for the triple (4, 1, 1) we have
, while the opposite inequality holds for (4, 4, 1). From now on, we restrict considerations to two numbers a, b. It was shown in the note [Citation1] that
, and conjectured that
for
, together with the opposite inequality for
. As reported in [Citation1], the conjecture was confirmed by Gord Sinnamon; his proof (communicated privately) is ingenious, but it involves some fairly heavy manipulation.
A more complete picture is obtained if at the same time we compare Mp with . Equality holds for p = 1, and it is easily verified that
(this is the harmonic mean), so equality also holds for p = – 1. The results in [Citation1] imply that
(with the logarithmic mean coming between these two quantities), suggesting that a similar inequality holds for
, though this was not explicitly stated as a conjecture.
Here we offer a simple unified treatment of both comparisons, based on the substitution that was used in [Citation1]. The results are as follows.
Theorem 1.
The inequality holds for
and for
. The opposite inequality holds for
and for p < 0.
Theorem 2.
The inequality holds for
and for
. The opposite inequality holds for
and for
.
Note first that if , then
and similarly for G and Mp, so it is sufficient to consider the pair
: henceforth the notation A, G, Mp applies to this pair. Now substitute
. Then
and
So, for example, the inequality stated above translates to
, which is obvious for p > 0. The inequality in Theorem 1 translates to
(1)
(1)
For both theorems, we will use the following Lemma, essentially an adaption of the “integrating factor” method to inequalities.
Lemma 3.
Let f be a function satisfying and
for
. Then
for t > 0. The reverse applies if
for t > 0.
Proof.
Let and
. Then
and
hence
Consequently and
are increasing. So for t > 0, we have
and
, hence
, so also
. The inequalities reverse if
. □
Proof of Theorem 1.
As we have seen, the substitution translates
to
(for all t), where
Since f is even, it is enough to consider t > 0. Then and
So and
If , then
, so
and
for all t. If
or p < 0, then
. So if
or p < 0, then
, and if
, then
. The statements follow, by the Lemma. □
Proof of Theorem 2.
The inequality translates to
(2)
(2)
(this inequality is perhaps of some interest in its own right). Let
Then and
where
If , then
, so
, hence
, so
for
. This implies (2) if
and the reverse of (2) if
. If
, then
, so
for
. This implies the reverse of (2) for
and (2) for
. □
It remains to compare and combine the inequalities in Theorems 1 and 2. There are five intervals to consider. For , we have
For , we have
In the other cases, we have either two upper bounds or two lower ones. We compare them. For this purpose, write . For
, we have
, so the better estimate is
, given by Theorem 1. (Recall that in this case we have the upper bound
).
For , we have
, by the weighted AM-GM inequality applied to
. So the better estimate is
, again from Theorem 1. (Also
).
For , we have again
, seen by writing
, where
. So the better estimate is
, given by Theorem 2. (Also
).
References
- Jameson, G. J. O., Mercer, P. R. (2019). The logarithmic mean revisited. Amer. Math. Monthly. 126(7): 641–645.