I am looking for a proof of weighted version of [the inequality](https://mathoverflow.net/questions/245092/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. $x\_1+...+x\_i \ge y\_1+...+y\_i$ 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) $



Weighted version of the Karamata Inequality
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. 


