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If $a_1\ge a_2\ge \cdots \ge a_n\ge 0$, $b_1\ge b_2\ge \cdots \ge b_n\ge 0$ with $\sum_{j=1}^kb_j=1$ for some $1\le k\le n$. Is it true that $2\sum_{j=1}^na_jb_j\le a_1+\frac{1}{k}\sum_{j=1}^na_j$?

The above question is denied.

Give a simple proof to a weaker version?

Under the same condtion, then $\sum_{j=1}^na_jb_j\le \max\{a_1,\frac{1}{k}\sum_{j=1}^na_j\}$.

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up vote 3 down vote accepted

No, it isn't true. Let $a_j=b_j=1$ for $1\leq j\leq n$. Then $\sum_{j=1}^1b_j=1$, hence $k=1$, thus we've satisfied the hypotheses. However, $$2\sum_{j=1}^na_jb_j=2n\geq1+n$$ with equality if and only if $n=1$.

EDIT: For the second case, let $a_j=1/10$ for all $j$, $b_1=1$ and $b_j=10$ for $j\geq 2$. Then we end up with $$n-0.9\geq n/10$$ with equality if and only if $n=1$ again.

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Thanks, I edited that. – Fischer Jan 15 '13 at 19:49
@Fischer: I've given you another counterexample. – Clayton Jan 15 '13 at 19:59
Oh, $b_j$ are in decreasing order. – Fischer Jan 15 '13 at 20:02
Well, I find a simple proof myself. – Fischer Jan 15 '13 at 21:36

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