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Let $f$ be a convex function from $R_{++}^n$ to $R$. If $f(x_i)\geq f(y_i)$, where $x_i,y_i$ in $R_{++}^n$, $i=1,2,...,n$. The question is: is the following inequality true: $$f(\sum_{i=1}^n a_i x_i)\geq f(\sum_{i=1}^n a_i y_i),$$ where $a_i\geq 0, \sum_{i=1}^n a_i=1$. If it is true, how to prove?

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What is $R^n_{++}$? –  Harald Hanche-Olsen Oct 29 '13 at 9:58

1 Answer 1

It seems the following.

This inequality is far away from the truth. For instance, let $f:\mathbb R\to \mathbb R$ such that $f(x)=x^2$ for each real $x$. Put $a_1=a_2=1/2$, $y_1=3,$ $y_2=-1$, $x_1=4$ and $x_2=-4$. Then $$f(x_1)=16>4=f(y_1), f(x_2)=9>1=f(y_2),$$ but $$f(a_1x_1+a_2x_2)=f(0)=0<1=f(1)=f(a_1y_1+a_2y_2).$$ The restriction of domain of $f$ to $\mathbb R_{++}$ is easily overcame by its shift.

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