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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 $i \in {1, 2,...,n}$

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    $\begingroup$ I don't think MO should be a place for people to keep posting conjectures $\endgroup$ – Yemon Choi Jul 26 '16 at 3:45
  • $\begingroup$ Dear @YemonChoi I think mathematics developed from conjecture and problem and observation $\endgroup$ – Oai Thanh Đào Jul 26 '16 at 4:11
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    $\begingroup$ I think, any question like "is the following proposition true?" may be called "a conjecture", and most MO questions are of this type. $\endgroup$ – Fedor Petrov Jul 26 '16 at 18:19
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    $\begingroup$ 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 '16 at 20:05
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    $\begingroup$ 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. $\endgroup$ – Fedor Petrov Jul 27 '16 at 10:50
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  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.

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  • $\begingroup$ 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=?$ $\endgroup$ – Oai Thanh Đào Jul 28 '16 at 11:11
  • $\begingroup$ $f=\max(x-2,0)$, other data are in the answer. $\endgroup$ – Fedor Petrov Jul 28 '16 at 11:37
  • $\begingroup$ Dear Dr. @FedorPetrov , the inequality (1) is hold ? with $x_1 \ge y_1$ and $x_2 \leq y_2$ $\endgroup$ – Oai Thanh Đào Jul 29 '16 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 $x_1\geqslant y_1\geqslant y_2\geqslant x_2$. $\endgroup$ – Fedor Petrov Jul 29 '16 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. $\endgroup$ – Oai Thanh Đào Jul 29 '16 at 16:58
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  1. I prove that the inequality (1) is hold with $x_1 \ge y_1$ and $y_2 \ge x_2$ as followings:

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

$$y'=\frac{f'(x)}{2}-\frac{1}{2}f'(\frac{x+x_2}{2})$$

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 $i \in {1, 2,...,n}$

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    $\begingroup$ Is your proposition 3 a theorem or conjecture or what? $\endgroup$ – Fedor Petrov Jul 30 '16 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. $\endgroup$ – Oai Thanh Đào Jul 30 '16 at 4:54
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    $\begingroup$ This new conjecture is correct, see the edit of my answer. $\endgroup$ – Fedor Petrov Aug 1 '16 at 19:11
  • $\begingroup$ DearDr. @FedorPetrov in case $x \ge y$, is the proof similarly? $\endgroup$ – Oai Thanh Đào Aug 2 '16 at 0:27
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    $\begingroup$ If $x>y$, we replace $x_i$ to $-x_i$, $y_i$ to $-y_i$ and thus reduce it to the case $x<y$. $\endgroup$ – Fedor Petrov Aug 2 '16 at 4:38

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