Let $x(a)\geq1$ and $y(b)\geq1$. I have a relation $x(a) \leq k(a,b)y(b)$ for all $k(a\geq b) \geq 1$ and $x(a)=y(b)$ when $k(a=b)=1$. Can we conclude that $x(a)\geq y(b)$ ?
Upper bound to lower bound
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algebra-precalculus
inequality
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0Parameter $k$ is actually affecting the inequality. I am trying to get $x>y$ in some condition. – 2011-08-26
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1Very confusing. $x,y$ scalars seems to mean they are constants. Second and third sentences seem to mean $x,y$ are functions of $k$. Last sentence, first half suggests they are functions of $k$, but second half makes more sense if they are numbers. Please rewrite question so it makes some sense. – 2011-08-27
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0I have rewritten..Thanks – 2011-08-27
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0I still don't understand the question. Are $x, y$ a function of $k$? – 2011-08-27
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0Why do you want this? Where come from $x$ and $y$? What is the context? – 2011-08-27
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0You have rewritten, but you have not fully engaged with my comment. If $x,y$ are functions of $k$, please say so. If they are not functions of $k$, please explain what's meant by "$x=y$ when $k=1$." – 2011-08-27
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0I am really sorry for not able to explain the problem in a nicer way. I have again modified...Hope it works..Thanks to all – 2011-08-28
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0shakera, your two hypothesis seem to be that $x(a)=y(a)$ for every $a$ and that $x(a)\le y(b)$ for every $a\ge b$ (since for positive $u$ and $v$, the hypothesis that $u\le kv$ for every $k\ge1$ is **equivalent** to the fact that $u\le v$). There is no way whatsoever these two hypothesis could imply $x(a)\ge y(b)$ for any $a\ne b$. This is a matter of logic alone (the inequality goes the wrong way), but here is a simple counterexample: choose any positive **decreasing** function $z$ and let $x(a)=y(a)=z(a)$ for every $a$. Can you check this is indeed a counterexample? – 2011-08-28
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0Maybe we could understand the question if you gave an example or two of numbers/functions/whatever $x,y,k$ (and $a$ and $b$?) satisfying the hypotheses of the question, and examples *not* satisfying the hypotheses. – 2011-08-28