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Question:

let postive numbers $n\ge 2$, such $a_{1},a_{2},\cdots,a_{n}>0$,and $b_{1},b_{2},\cdots,b_{n}\in R$,and such $$\sum_{1\le i\le n}a_{i}b_{i}=0,|b_{1}|+|b_{2}|+\cdots+|b_{n}|\neq 0$$ show that $$\sum_{1\le i<j\le n}b_{i}b_{j}<0$$

My idea: since $$|b_{1}|+|b_{2}|+\cdots+|b_{n}|\neq 0$$ so $b_{i}$ is not all zero and $$\sum_{1\le i<j\le n}b_{i}b_{j}=\sum_{i=1}^{n-1}a_{i}\left(\sum_{j=i+1}^{n}b_{j}\right)=\sum_{j=2}^{n}b_{j}\left(\sum_{i=1}^{j-1}b_{i}\right)$$ so we must prove follow inequality $$\sum_{j=2}^{n}b_{j}\left(b_{1}+b_{2}+\cdots+b_{j-1}\right)<0$$

then I can't Continue.Thank you

math110
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  • are the two summations, double summations? I mean, is this correct: $\sum {1\le i \lt j \le n} a_ib_j = \sum{1\le i \le n} \sum_{n\ge j\gt i} a_ib_j$ – tpb261 Jun 02 '14 at 05:43
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    For $n=3$ with $a_1=a_2=a_3=1$ and $b_1=10,b_2=1,b_3=-1/2$ the sum $\sum_{i<j}a_ib_j=11+1(-1/2)+1(-1/2)=0$ and yet the other sum $\sum_{i<j}b_ib_j=101+10(-1/2)+1(-1/2)=+4.5.$ Or am I interpreting the summation range wrong? Usually if $i<j$ appears below a sum it means only pairs for which $i<j$ enter into the sum. – coffeemath Jun 02 '14 at 07:03
  • oh.!@coffeemath,Nice conutexmple! – math110 Jun 02 '14 at 07:16
  • and Now I have edit it,because My frends tell me is wrong before problem, – math110 Jun 02 '14 at 07:24
  • @math110 Yes, now that the indexing on the assumption is over the products $a_ib_i$ it looks more likely, since at least this way all the a's and all the b's enter into the assumption. Maybe a difficult inequality to show, +1 on question. – coffeemath Jun 02 '14 at 08:06

2 Answers2

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This problem is totally wrong. Let $$n=3,\quad a_1=a_2=b_1=b_2=1,\quad a_3=5,b_3=-2/5,$$ then $$\sum_{i=1}^3a_ib_i=2+5\times(-2/5)=0.$$ However, $$b_1b_2+b_1b_3+b_2b_3=1-4/5=1/5>0.$$ The counterexamples can be constructed easily by substitute $b_3=-\dfrac{a_1b_1+a_2b_2}{a_3}$ into the inequality when $n=3$. We have $$b_1b_2+b_1b_3+b_2b_3=\frac{a_3b_1b_2-(b_1+b_2)(a_1b_1+a_2b_2)}{a_3}.$$Suppose $b_1,b_2>0$, one can simply let $a_3$ go to infinity, then $b_1b_2+b_1b_3+b_2b_3>0$.

Nirvanacs
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$$ 0 = \sum_{1\le i \le n} a_ib_i $$ $$ \implies 0 = (\sum_{1\le i \le n} a_ib_i)^2 = \sum_{1\le i\le n} (a_ib_i)^2 + 2(\sum_{1\le i \le n} \sum_{i\lt j \le n} a_ia_jb_ib_j) $$

Since, $a_i>0, \forall i$ divide both sides of the above by $(2*MAX(a_i))^2)$ and then the required inequality can be got.

$$ \sum_{1\le i \le n} \sum_{i\lt j \le n} a_ia_jb_ib_j = -\frac{1}{2} \sum_{1\le i\le n} (a_ib_i)^2 <0 $$ $$ \implies \sum_{1\le i \le n} \sum_{i\lt j \le n} b_ib_j <0 $$

tpb261
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  • -1 I don't understand why you can divide both sides by $2A^2$. For example, we may know that $ 531(-1) + 3111 < 0$, but that doesn't imply that $ 1 * (-1) + 1*1 < 0$. You cannot change the positive and negative values by varying amounts, and still expect the equality to hold true. – Calvin Lin Jun 04 '14 at 04:30
  • I's divide by the $MAX^2(5,3,3,1) = 25$ – tpb261 Jun 04 '14 at 08:50
  • You're missing the point. Dividing by 25 doesn't help. When you delete the $a_i$ (represented by the first 2 terms,) you're making the negative value much smaller, but the positive value isn't made much smaller. – Calvin Lin Jun 04 '14 at 18:58
  • @CalvinLin Yeah.. you're right... I guess I'll leave the answer till a proper one comes up and then delete it. It'll be a clue of what not to do. – tpb261 Jun 05 '14 at 02:46