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 December 12th, 2012, 06:49 AM #1 Newbie   Joined: Dec 2012 Posts: 1 Thanks: 0 Can you prove this inequality? Hi guys, I was wondering if someone can help me to solve this inequality, which is as follows: Given $2p$ non-negative numbers $b_1,b_2,...,b_p,c_1,c_2,c_p$ such that $b_1+b_2+...+b_p=1$ and $b_1c_1\geq b_kc_k$ for all $k=1,2,...,p$, show that following holds: $b_1c_1\geq \frac{1}{\frac{1}{c_1}+\frac{1}{c_2}+...+\frac{1}{ c_p}}$. I tried to solve this with AM-HM inequality and few other ways, but without success. Only thing I can prove is for special case $p=2$. I've checked this programmatically (in a few ways) and haven't found any counterexample, so I think it should be correct. I proposed this by myself, haven't found anywhere else on the internet. Does anyone know solution? Or anyone can give me some suggestion? Thanks in advance!
 December 12th, 2012, 08:56 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Can you prove this inequality? Hmm... the first approach that springs to mind is to alter the statement to $b_1+...+b_{p-1}<1$, $b_1c_1\ge b_kc_k$ for $k=1,\ldots,p-1$ such that $b_1c_1\ge\frac{1}{1/c_1+...+1/c_{p-1}+1/c}$ for any $c\le b_1c_1/(1-b_1-...-b_{p-1}).$ This way hopefully it is easier to use induction, since the last term is implicit so you can re-use earlier cases.
 December 19th, 2012, 06:33 PM #3 Senior Member   Joined: Dec 2012 Posts: 148 Thanks: 0 Re: Can you prove this inequality? $pb_{1} c_{1} \geq \sum_{k=1}^{p} b_{k} c_{k} \geq \sum_{k=1}^{p} c_{k}$ $b_{1} c_{1} \geq \frac{ \sum_{k=1}^{p} c_{k} }{p} \geq \frac{p}{\sum_{k=1}^{p} \frac{1}{c_{k}} } \geq \frac{1}{ \sum_{k=1}^{p} \frac{1}{c_{k}}}$
December 25th, 2012, 10:51 PM   #4
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Re: Can you prove this inequality?

Quote:
 Originally Posted by tahir.iman $pb_{1} c_{1} \geq \sum_{k=1}^{p} b_{k} c_{k} \geq \sum_{k=1}^{p} c_{k}$ $b_{1} c_{1} \geq \frac{ \sum_{k=1}^{p} c_{k} }{p} \geq \frac{p}{\sum_{k=1}^{p} \frac{1}{c_{k}} } \geq \frac{1}{ \sum_{k=1}^{p} \frac{1}{c_{k}}}$
How did you prove $\sum_{k=1}^{p} b_{k} c_{k} \geq \sum_{k=1}^{p} c_{k}$ ?

December 25th, 2012, 11:01 PM   #5
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Re: Can you prove this inequality?

Quote:
Originally Posted by golomorf
Quote:
 Originally Posted by tahir.iman $pb_{1} c_{1} \geq \sum_{k=1}^{p} b_{k} c_{k} \geq \sum_{k=1}^{p} c_{k}$ $b_{1} c_{1} \geq \frac{ \sum_{k=1}^{p} c_{k} }{p} \geq \frac{p}{\sum_{k=1}^{p} \frac{1}{c_{k}} } \geq \frac{1}{ \sum_{k=1}^{p} \frac{1}{c_{k}}}$
How did you prove $\sum_{k=1}^{p} b_{k} c_{k} \geq \sum_{k=1}^{p} c_{k}$ ?
Oh, i have mistake there, i am on it.

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