Answer by Especially Lime for Prove that $x_1+x_2+\cdots+x_L\geq...
You can use the fact that$$(1-a_1)x_1+\cdots+(1-a_L)x_L\geq (1-a_1+\cdots+1-a_L)x_L$$together with $(1-a_1+\cdots+1-a_L)-a_{L+1}-\cdots-a_n=0$.
View ArticleProve that $x_1+x_2+\cdots+x_L\geq a_1x_1+a_2x_2+\cdots+a_nx_n$
Given $x=[x_1\quad x_2\quad \ldots \quad x_n],$ such that $x_1\geq x_2\geq\cdots\geq x_n$. Prove the following inequality:\begin{equation}x_1+x_2+\cdots+x_L\geq...
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