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I hope the following kind of inequality holds: let $a_i,b_i\in R$ with $b_i>0$, $\sum _{i=1}^mt_i=1$ with $t_i>0$, then $$\frac{t_1a_1+\cdot+t_ka_k}{t_1b_1+\cdot+t_kb_k}\le\frac{a_1}{b_1}+\cdot+\frac{a_k}{b_k}.$$

I am not sure whether it is true. Does someone know some inequalities like above? Many thanks!

Note: I have changed the inequality. The following $$\frac{t_1a_1+\cdot+t_ka_k}{t_1b_1+\cdot+t_kb_k}\le \max(\frac{a_1}{b_1},\cdots,\frac{a_k}{b_k}).$$ holds. The matter is $a_i (i=1,\cdots,k) $ are not necessary positive.

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closed as not a real question by Yemon Choi, Andres Caicedo, Pietro Majer, Michael Renardy, quid Dec 11 '12 at 17:07

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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Your statement cannot be true. You need at very least to ask whether there is $C>0$ such that [...] and to impose that also $a_i>0$ (otherwise you could simply obtain the reverse inequality simply by flipping the sign of $a_i$). –  Delio Mugnolo Dec 10 '12 at 11:09
    
...or $a_i\ge 0$, of course. –  Delio Mugnolo Dec 10 '12 at 11:11
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Suppose Delio's comments were imposed, i.e. you look only at $a_i \geq 0$ (I see you have $C>0$ now.). As it is now you've "given too much" to be a meaningful inequality, because you've given the $a_i , b_i,$ and $t_i$. This means that the right hand side is fully determined, as is everything on the left hand side except for C. Also both sides are nonnegative. Thus, the question reduces to asking if the left hand side is nonzero, whether there is a positive number $C \leq $ to the quotient of two positive numbers, which is always a positive number so yes. –  Robert L. Simione II Dec 10 '12 at 11:52
    
My guess is that a more meaningful inequality would have $C$ depend only on the $t_i$, but that's for you to decide. –  Robert L. Simione II Dec 10 '12 at 11:53
    
Thank you for your comments. –  ljjpfx Dec 11 '12 at 11:00

2 Answers 2

Write $x_i = a_i/b_i$, $y_i= t_i b_i$. Your inequality then reduces to the question of if there is some $C$ such that sometimes $$ C \sum_i x_i y_i \le \sum_i x_i \cdot \sum_i y_i, $$ and if so, under what conditions.

One answer to this question goes by the name of Chebyshev's inequality in math olympiad circles; it's sometimes called Chebyshev's order inequality or Chebyshev's sum inequality to distinguish it from the (apparently unrelated?) result in probability theory. It can be found in Hardy–Littlewood–Pólya, Steele's book The Cauchy–Schwarz Masterclass, and various places on the internet.

In brief, $C=n$, and the condition is that the $x_i$ and $y_i$ have to be oppositely sorted (meaning that we can permute the indices such that $x_1 \le \dotsb \le x_n$ and $y_1 \ge \dotsb \ge y_n$). The $y_i$ don't have to be non-negative, but the sorting condition is crucial: if they are similarly sorted, then the inequality flips its direction!

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Is there any reason this should be true with k = 2? I see, after some algebra, a ratio of cubics being compared to t, with 0 < t < 1.

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