# Weighted version of the Karamata Inequality

I am looking for a proof of weighted version of the 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. $x_1+...+x_i \ge y_1+...+y_i$ and $x_{i+1}+...+x_n \le y_{i+1}+...+y_n$ for $i=1,...,n-1$

3. Let $\lambda_i >0$ and $\sum_1^{n} \lambda_i =1$

Then show that:

$$\lambda_1f(x_1)+\lambda_2f(x_2)+...+\lambda_nf(x_n)-f(\lambda_1x_1+\lambda_2x_2+...+\lambda_nx_n) \ge \lambda_1f(y_1)+\lambda_2f(y_2)+...+\lambda_nf(y_n)-f(\lambda_1y_1+\lambda_2y_2+...+\lambda_ny_n)$$

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

Remark:

If $\lambda_1x_1+\lambda_2x_2+....+\lambda_nx_n=\lambda_1y_1+\lambda_2y_2+....+\lambda_ny_n$

Then we have the weighted version of the Karamata inequality:

$\lambda_1f(x_1)+\lambda_2f(x_2)+...+\lambda_nf(x_n) \ge \lambda_1f(y_1)+\lambda_2f(y_2)+...+\lambda_nf(y_n)$

• In weighted Karamata you need to include the weights into the majorization condition. Jan 1, 2021 at 4:58

This is false by some stupid reasons. Say, if $\lambda_1=\lambda_n=0$, then what we have to prove does not depend on $x_1,x_n,y_1,y_n$, but changing these four variables with fixed other variables we may easily satisfy all the conditions.
• Dear Dr @FedorPetrov . Thank to You, but I don't think so, because when $\lambda_1=\lambda_n=0$ then $x_1, x_n$ and $y_1, y_n$ be removed in two sides. Aug 27, 2016 at 10:53