We give a necessary and sufficient mean condition for the quotient of two Jensen functionals and define a new class
It is said that the mean
It is also well known that
An easy task is to construct intermediate means related to two given means
The problem is more difficult if we have to decide whether the given mean is intermediate or not. For example, the relation
Also,
An inverse problem is to find best possible approximation of a given mean
Since
Finally, Lin showed in [
Numerous similar results have been obtained recently. For example, an approximation of Seiffert’s mean by the class
In this paper we will give best possible approximations for a whole variety of elementary means (
Let
It turns out that the expression
Let
As an illustration, consider the function
Since
Moreover, it is evident that
We will give in the sequel a complete answer to the above question concerning the means
Those means are obviously symmetric and homogeneous of order one.
As a consequence we obtain some new intermediate mean values; for instance, we show that the inequalities
We prove firstly the following
Let
In the same way, for arbitrary
A generalization of the above assertion is the next.
Let
It should be noted that the relation
Our results concerning the means
For the class of means The means there is a number there is a number there is no finite
The above estimations are best possible.
We prove firstly the necessity of the condition (
Since
From the other hand, due to l’Hospital’s rule we obtain
Suppose now that (
But,
Therefore, by (
We will give a proof of this assertion by induction on
By Remark
Next, it is not difficult to check the identity
Therefore, by induction hypothesis and Remark
The inequality
For the proof of necessity, put
It is evident from (
Because the above inequality does not depend on
For an arbitrary probability law
We will prove a general assertion of this type. Namely, for an arbitrary positive sequence
For
This assertion follows applying the result from [
For
Putting there
Since
The part (1) of Theorem
A general way to prove the rest of Theorem
Since
Therefore, we have to compare some one-variable inequalities and to check their validness for each
For example, we will prove that the inequality
Since
By the above formulae, this is equivalent to the assertion that the inequality
We will prove that the power series expansion of
Since
Hence,
Now, one can easily prove (by induction, e.g.) that
Therefore,
Similarly, we will prove that the inequality
As before, it is enough to consider the expression
It is not difficult to check the identity
Hence by (
Therefore
By monotonicity it follows that
For
Hence,
From the other hand,
Examining the function
Since
A calculation gives
Note also that
Therefore, applying the assertion from the part 1, we get
Finally, we give a detailed proof of the part 7.
We have to prove that
Therefore, by the transformation given above, we get
Further, we have to show that
Indeed, since
As for the part 8, applying the above transformation we obtain
Since for
The rest of the proof is straightforward.
The author is indebted to the referees for valuable suggestions.