# Upper bound to lower bound

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 $x(a)\geq y(b)$ ?

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Parameter $k$ is actually affecting the inequality. I am trying to get $x>y$ in some condition. – shakera Aug 26 '11 at 18:06
Very 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. – Gerry Myerson Aug 27 '11 at 1:06
I have rewritten..Thanks – shakera Aug 27 '11 at 9:07
I still don't understand the question. Are $x, y$ a function of $k$? – Qiaochu Yuan Aug 27 '11 at 16:45
Why do you want this? Where come from $x$ and $y$? What is the context? – leo Aug 27 '11 at 18:12

## 1 Answer

No, it most definitely does not imply that. Indeed, try solving for both $x< k y$ and $x \le y$ at the same time for $x \ge 1 \land y \ge 1$.

For $k=2$, $x = 3$ and $y = 5$ clearly satisfy both: $3 < 2 \times 5$ and $3 \le 5$.

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Thanks for your answer. – shakera Aug 26 '11 at 18:00
This example does not satisfy that when $k=1$, $x=y$. – shakera Aug 26 '11 at 18:15
Of course it does not, but your question was specifically formulated for $k>1$. – Sasha Aug 26 '11 at 18:16
It should also satisfy for $k=1$, $x=y$. – shakera Aug 26 '11 at 18:53