Fist I observe function $f(x)=x^2$ in the figure as following

enter image description here

I found that when $x_1 \ge y_1$ and $x_2 \le y_2$ $\Rightarrow$ $AB \ge CD$ $\Rightarrow$ $$\frac{f(x_1)+f(x_2)}{2}-f(\frac{x_1+x_2}{2}) \ge \frac{f(y_1)+f(y_2)}{2}-f(\frac{y_1+y_2}{2})$$ (1).

Note that: When $x_1+x_2=y_1+y_2$ the inequality (1) is Karamata inequality (with case n=2)

$$f(x_1)+f(x_2) \ge f(y_1)+f(y_2)$$

From above observation, I am looking for a proof of a conjecture generalization of Karamata inequality as following:

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:

  1. $x_1 \ge x_2 \ge x_3...\ge x_n,$ and $y_1 \ge y_2 \ge y_3...\ge y_n$

2.x_1+x_2+...+x_i \ge y_1+y_2+...+y_i for i=1,...,n-1 and

3. x_n \le y_n

2'. $x_1+...+x_i \ge y_1+...+y_i$ and $x_{i+1}+...+x_n \le y_{i+1}+...+y_n$ for $i=1,...,n-1$ then

$$\frac{f(x_1)+f(x_2)+...+f(x_n)}{n}-f(\frac{x_1+x_2+...+x_n}{n}) \ge \frac{f(y_1)+f(y_2)+...+f(y_n)}{n}-f(\frac{y_1+y_2+...+y_n}{n}) $$

The inequality holds with equality if and only if $x_i=y_i$ for all $i \in {1, 2,...,n}$

  • 6
    $\begingroup$ I don't think MO should be a place for people to keep posting conjectures $\endgroup$
    – Yemon Choi
    Jul 26, 2016 at 3:45
  • $\begingroup$ Dear @YemonChoi I think mathematics developed from conjecture and problem and observation $\endgroup$ Jul 26, 2016 at 4:11
  • 7
    $\begingroup$ I think, any question like "is the following proposition true?" may be called "a conjecture", and most MO questions are of this type. $\endgroup$ Jul 26, 2016 at 18:19
  • 3
    $\begingroup$ I think both Yemon Choi and Fedor Petrov are right. -- It's just that questions should be formulated as such. As you have done this in your last edits, it is no longer unclear what you're asking, and hence I have voted to reopen. $\endgroup$
    – Stefan Kohl
    Jul 26, 2016 at 20:05
  • 1
    $\begingroup$ Inequality $\sum f(a_i)\geq \sum f(b_i)$ for any convex function holds if and only if it holds by Karamata, that is, if and only if the multiset $\{a_i\}$ majorates the multiset $\{b_i\}$. In our situation $\{a_i\}$ is the set consisting from $x_1,\dots,x_n$ and $n$ times $(\sum y_i)/n$, another multiset $\{b_i\}$ consists of $y_1,\dots,y_n$ and $n$ times $(\sum x_i)/n$. You may try to prove this by considering many cases or by procedures like 'bring to $x$'s together' or 'move two $y$'s apart', when they do not violate your inequalities. $\endgroup$ Jul 27, 2016 at 10:50

2 Answers 2

  1. Conditions (1,2,3) are not enough for the inequality to hold. Take $n=3$, $x_1=x_2=3,x_3=0$, $y_1=3,y_2=y_3=0$. Then we need the multiset $(3,3,0,1,1,1)$ (all three $x$'s and 3 times mean of $y$'s) to majorate $(3,0,0,2,2,2)$. But four largest elements of the first multiset have sum $3+3+1+1=8$, while in the second it is $3+2+2+2=9$. So, the claim does not hold in full generality. To be more explicit, we get an opposite inequality for $f(x)=\max(x-2,0)$.

  2. Conditions (1) and

(2') $x_1+...+x_i \geqslant y_1+.....+y_i$ and $x_{i+1}+...+x_n \leqslant y_{i+1}+...+y_n$ for $i=1,....,n-1$

are enough. We prove it by verifying that the multiset of $2n$ numbers $A:=\{x_1,\dots,x_n,y,y,\dots,y\}$ majorates the multiset $B:=\{y_1,\dots,y_n,x,x,\dots,x\}$, where $x=\frac1n \sum x_i$, $y=\frac1n \sum y_i$. Without loss of generality $x\leqslant y$, else change signs of all $x$'s and $y$'s. We have to check that the sum $w_m$ of $m$ largest elements of $B$ does not exceed the sum of $m$ largest elements of $A$. Let $m$ largest elements of $B$ be $y_1,\dots,y_s$ and $(m-s)$ times $x$. Consider two cases.

1-st case. $s\leqslant n-1$. Then $w_m=y_1+\dots+y_s+(m-s)x\leqslant x_1+\dots+x_s+(m-s)y$.

2-nd case. $s=n$. Then $w_m=y_1+\dots+y_n+(m-n)x\leqslant n\cdot y+x_1+\dots+x_{m-n}$.

In both cases we found $m$ elements of $A$ with a sum at least $w_m$, as desired.

  • $\begingroup$ Thank to dear @FedorPetrov Please let me an example: $f(x)=?$, $n=?$ $x_1, x_2,....,x_n= ?$ and $ y_1=?, y_2=?,...y_n=?$ $\endgroup$ Jul 28, 2016 at 11:11
  • $\begingroup$ $f=\max(x-2,0)$, other data are in the answer. $\endgroup$ Jul 28, 2016 at 11:37
  • $\begingroup$ Dear Dr. @FedorPetrov , the inequality (1) is hold ? with $x_1 \ge y_1$ and $x_2 \leq y_2$ $\endgroup$ Jul 29, 2016 at 14:55
  • $\begingroup$ @OaiThanhĐào yes, the expression $h(x_1,x_2)$ in LHS is increasing in $x_1$ and decreasing in $x_2$ (when $x_1\geqslant x_2$), therefore $h(x_1,x_2)\geqslant h(y_1,x_2)\geqslant h(y_1,y_2)$ if $x_1\geqslant y_1\geqslant y_2\geqslant x_2$. $\endgroup$ Jul 29, 2016 at 15:07
  • $\begingroup$ Dear Dr. @FedorPetrov could you give for condition of xi, yi such that the inequality hold. Please read my ask below. $\endgroup$ Jul 29, 2016 at 16:58
  1. I prove that the inequality (1) is hold with $x_1 \ge y_1$ and $y_2 \ge x_2$ as followings:

$$y(x)=\frac{f(x)+f(x_2)}{2}-f(\frac{x+x_2}{2})- \frac{f(y_1)+f(y_2)}{2}+f(\frac{y_1+y_2}{2})$$


Because $f'' \ge 0$ so $y'(x) \ge y'(x_2)=0$ because $x \ge x_2$, because $x_1 \ge y_1$ so $y(x_1) \ge y(y_1)$, therefor

$$y(x_1) \ge y(y_1)=\frac{f(y_1)+f(x_2)}{2}-f(\frac{y_1+x_2}{2})- \frac{f(y_1)+f(y_2)}{2}+f(\frac{y_1+y_2}{2})=\frac{f(x_2)}{2}-f(\frac{y_1+x_2}{2})- \frac{f(y_2)}{2}+f(\frac{y_1+y_2}{2}) $$

Let $$g(t)=\frac{f(x_2)}{2}-f(\frac{t+x_2}{2})- \frac{f(y_2)}{2}+f(\frac{t+y_2}{2}) $$ with $t \ge y_2$, we have:

$$g'(t)=-\frac{1}{2}f'(\frac{t+x_2}{2})+\frac{1}{2}f'(\frac{t+y_2}{2}) $$

Because $f'' \ge 0$ and $y_2 \ge x_2$ so $g'(t) \ge 0$ so

$g(y_1) \ge g(y_2)=\frac{f(x_2)}{2}-f(\frac{y_2+x_2}{2})- \frac{f(y_2)}{2}+f(\frac{y_2+y_2}{2})=\frac{f(x_2)+f(y_2)}{2}-f(\frac{y_2+x_2}{2}) \ge 0$ because $f'' \ge 0$

  1. Remark with the same proof, I can prove the inequality (1) with weights $\lambda_i$ holds.

Let $x_1 \ge y_1$ and $x_2 \le y_2$, $\lambda_i >0 $ and $\lambda_1+\lambda_2=1$ we have:

$$\lambda_1f(x_1)+\lambda_2f(x_2)-f(\lambda_1x_1+\lambda_2x_2) \ge \lambda_1f(y_1)+\lambda_2f(y_2)-f(\lambda_1y_1+\lambda_2y_2)$$ (2)

  1. Now 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:

(i). $x_1 \ge x_2 \ge x_3...\ge x_n,$ and $y_1 \ge y_2 \ge y_3...\ge y_n$

(ii). $x_1+...+x_i \ge y_1+...+y_i$ and $x_{i+1}+...+x_n \le y_{i+1}+...+y_n$ for $i=1,...,n-1$

$$\frac{f(x_1)+f(x_2)+...+f(x_n)}{n}-f(\frac{x_1+x_2+...+x_n}{n}) \ge \frac{f(y_1)+f(y_2)+...+f(y_n)}{n}-f(\frac{y_1+y_2+...+y_n}{n}) $$

The inequality holds with equality if and only if $x_i=y_i$ for all $i \in {1, 2,...,n}$

  • 1
    $\begingroup$ Is your proposition 3 a theorem or conjecture or what? $\endgroup$ Jul 30, 2016 at 4:38
  • $\begingroup$ Proposition 3 is a conjecture. Some days recenterly I think how I can generalization of (1). So I give stronger condition of $x_i$ and $y_i$ in conjecture 3. I am looking for the proof. $\endgroup$ Jul 30, 2016 at 4:54
  • 1
    $\begingroup$ This new conjecture is correct, see the edit of my answer. $\endgroup$ Aug 1, 2016 at 19:11
  • $\begingroup$ DearDr. @FedorPetrov in case $x \ge y$, is the proof similarly? $\endgroup$ Aug 2, 2016 at 0:27
  • 1
    $\begingroup$ If $x>y$, we replace $x_i$ to $-x_i$, $y_i$ to $-y_i$ and thus reduce it to the case $x<y$. $\endgroup$ Aug 2, 2016 at 4:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.