Dear Miodrag Mateljević,

Yes, it is pretty.

The literature is full of similar nice inequalities.

Especially that one related to (majorization ) as in the attached papers.

The natural question: is the relation works for the convex functions of several variables? under a suitable definition of sums in the convex domains.

Best regards

Dear prof Issam Kaddoura,

Thanks for comments, question and for enclosed paper.

It seems that as you mentioned

''The natural question: is the relation works for the convex functions of several variables? under a suitable definition of sums in the convex domains. ''

In order to outline a vivid interpretation of MP-inequality we appeal on our paper

[1] THE TEACHING OF MATHEMATICS

2010, Vol. XIII, 1, pp. 1–16

A CONTRIBUTION TO THE DEVELOPMENT OF FUNCTIONAL

THINKING RELATED TO CONVEXITY (by Miodrag Mateljevic and Marek Svetlik).

See Abstract of [1] below.

Now we outline our interpretation.

Take n-copies of a vessel (say vessel V-1)and cut them on the heights $h_1$, $h_2$, ... and $h_n$ to produce vessels $V_1$, ..., $V_n$.

By cutting part of the bases of these vessels, form a new container V-2 (we leave details here). Suppose that the height of vessel V-1 is $h_0=h_1+... +h_n$

If at the same time we begin to fill both containers, and if the height filling function for the vessel V-1 is convex (the level of the liquid changes accelerating)

the vessel V-1 will first be filled. This an interpretation of MP-inequality (we leave details here).

Abstract of [1].

When a liquid (water) flows into a vessel at the constant inflow rate,

then the height filling function is convex or concave depending on the way how the

level of the liquid changes. When the level changes accelerating or slowing down, the

function is convex or concave, respectively. This vivid interpretation holds in general,

namely we prove that given a strictly increasing convex (concave) continuous function,

then there exists a vessel (container) such that its height filling function is equal to the given

function. (A fact that seems to be new.)

We also hope that our paper could exemplify the case of a research project to be

assigned to excellent students.

We give a few comments:

Any local minimum of a convex function is also a global minimum. A strictly convex function will have at most one global minimum.

"If f is strictly convex in a convex set, show it has no more than 1 minimum". Math StackExchange. 21 Mar 2013. Retrieved 14 May 2016.

See https://en.wikipedia.org/wiki/Convex_function .

Hence it seems that if f is strictly convex on an interval (a,b), then

there are three possibility

i) f is increasing on (a,b)

ii) f is decreasing on (a,b)

iii) there a point c in (a,b) such that f is decreasing on (a,c) and

f is increasing on (c,b)

Note that in particular from MP-inequality follows if f is a convex function of one real variable, and f(0) = 0, then f is superadditive on the positive reals.

Can we improve MP-inequality (or Karamata-inequality ) in some of this case ?

Dear Miodrag Mateljević,

You said .

In which sense?

Best regards

Karamata-inequality .

Let $I$ be an interval of the real line and let $f$ denote a real-valued, convex function defined on I. If $x_1, . . . , x_n$ and $y_1, . . . , y_n$ are numbers in $I$ such that $(x_1, . . . , x_n)$ majorizes $(y_1, . . . , y_n)$, then

(1) $f(x_1)+\cdots+f(x_n) \geq f(y_1)+\cdots+f(y_n)$.

For example, If the convex function $f$ is non-decreasing, then that the equality in definition of majorization can be relaxed to

(a) $x_1+\cdots+x_n\geq y_1+\cdots+y_n$.

Set $k_f(x,y)= \frac{f(y)-f(x)}{y-x}$. If f is increasing on (a,b), then $k_f(x,y)>0$ for $x\neq y$. Can we use it?

Dear Prof. Miodrag Mateljević

Yes we can use it and we can prove Karamata (majorization ) inequality under these conditions

In the literature we can find:

Let $I$ be an interval of the real line and let $f$ denote a real-valued, convex function defined on I. If $x_1, . . . , x_n$ and $y_1, . . . , y_n$ are numbers in $I$ such that $(x_1, . . . , x_n)$ majorizes $(y_1, . . . , y_n)$, then\\

(1) $f(x_1)+\cdots+f(x_n) \ge f(y_1)+\cdots+f(y_n)$.

Here majorization means that

(2) $x_1+\cdots+x_n=y_1+\cdots+y_n$\\

and, after relabeling the numbers $x_1, . . . , x_n$ and $y_1, . . . , y_n$, respectively, in decreasing order, i.e.,\\

(3) $x_1\ge x_2\ge\cdots\ge x_n$ and $y_1\ge y_2\ge\cdots\ge y_n$,

we have

(4) $x_1+\cdots+x_i\ge y_1+\cdots+y_i$ for all $i \in \{1, . . . , n -1\}$.

At first glance it seems that the sentence ''after relabeling the numbers $x_1, . . . , x_n$ and $y_1, . . . , y_n$, respectively, in decreasing order ...'' should be clarified. If I remember well sometimes in the literature the condition (3) is omitted. Perhaps someone has some comments about it.

The condition (3) is important. In some places the author suppose that the tuples are decreasing for example given in THE TEACHING OF MATHEMATICS and this is they just taking for simplecity .

It is one old, but nice inequality of Mihailo Petrović (1868-1943), Serbian mathematician, founder of Belgrade's school of Analysis.

In connection with the above discussion a pdf.file is enclosed, which is very rough version of work in progress.

PS. Marcia Ricci Pinheiro asked to send corresponding materials.

In some situation we can improve MP-inequality (Mihailo Petrović ) .

Set $s_k= x_1 + ... + x_k$, $k=1,2,...,n$, and $s_0=0$.

For function $f$ and given points $x_1, ...,x_n$ set $x_0=0$, $I_k=(s_k,s_{k+1}]$.

Define $\tilde{f}$ by induction $\tilde{f} (x)=\tilde{f} (x - s_k) - \tilde{f} (0) +\tilde{f} (s_k)$ for $x\in I_k$.

(i) If $f : [0,\infty) \rightarrow \mathbb{R}$ is a convex function, then $R= f- \tilde{f}$ is not decreasing

(ii) If $f$ in addition has a minimum at $t_m= t_{min}$, then we can improve MP-inequality.

For example, if $0< s_1< s_2 < t_m < s_3 < s_4$ and $x_k < t_m$,$k=1,2,3,4$, then

$R(s_4)\geq R(t_m)+ |f(s_4) - f(t_m)| + |f(0) - f(s_3 - t_m)| + |f(0) - f(s_3)|$.