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 a_1x_1+a_2x_2+\cdots+a_nx_n
\end{equation}
for all $L=1,2,\ldots,n$ and $a_1+a_2+\cdots+a_n=L$ and $0\leq a_i\leq 1$.
My attempt: I understand the idea behind, on LHS we take largest $L$ elements, while on the LHS we are taking weighted sum, which will be less. But how to prove it mathematically?
\begin{equation}
(1-a_1)x_1+(1-a_2)x_2+\cdots+(1-a_L)x_L\geq a_{L+1}x_{L+1}+\cdots+a_nx_n
\end{equation}
Equality achieved only when $a_i=1$ for $i=1,2,\ldots,L$ and $a_i=0$ for $i=L+1,L+2,\ldots,n$, then LHS=RHS=$0$.
Best Answer
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 $(1-a_1+\cdots+1-a_L)-a_{L+1}-\cdots-a_n=0$.