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I'm currently working on a math problem that involves proving an inequality involving real numbers $a$, $b$, and $c$. The given inequality is $a + c \epsilon \leq b$, for all $\epsilon$ in the open interval $(0,1)$.

The complete math question: Let $a$, $b$, and $c$ be real numbers (constants). Given any positive number $\epsilon \in (0,1)$, if the inequality $a + c \epsilon \leq b$ holds for all such $\epsilon$, prove that $a \leq b$.

I'm looking for assistance in proving that $a \leq b$ based on this inequality.I'm trying using the contradiction method to prove this. I can get $a-b \gt 0$,so I want to get an contradiction by $b-a \geq c \epsilon$.Maybe,we can give the former inequality some conditon,then $a-b \gt \epsilon$.What I want is a contradiction about the order of $a$ and $b$.

I believe this problem may require some creative thinking and careful analysis of the given conditions. Any insights, suggestions, or step-by-step solutions would be greatly appreciated! Thank you in advance for your help.

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  • $\begingroup$ I think you are right.Thank you for your help. $\endgroup$
    – Leonardo Z
    Apr 14, 2023 at 2:22

2 Answers 2

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If $c \ge 0$ then $b \ge a+\frac12c \ge a$ and you're done.

If $c < 0$, take $c' = -c > 0$, so that $a - c'\varepsilon \le b$. If $\lnot(a\le b),$ then $a - b > 0$ so $a - b < c'\varepsilon$. A few more algebraic manipulations (keeping in mind that everything is now positive) and you get that $\frac1{\varepsilon} < \frac{c'}{a-b}$. Take $\varepsilon = 1/n$ for $n \in \mathbb N$. Then $n < \frac{c'}{a-b}$ for all $n\in \mathbb N$ which contradicts the archimedian property of the reals.

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We consider the following cases:

  1. $c=0$. Then $a+c\varepsilon\leq b\implies a\leq b$ and we are done.
  2. $c>0$. Then $a+c\varepsilon\leq b\implies\varepsilon\leq\frac{b-a}{c}$. Now since $\varepsilon \in (0,1)$, we have that $\varepsilon^k<\varepsilon^{k-1}<\dots<\varepsilon^2<\varepsilon\leq\frac{b-a}{c}, \forall k \in\mathbb{N}$, that is, $$\varepsilon\leq\frac{b-a}{c}\\ \varepsilon^2\leq\frac{b-a}{c}\\ \vdots\\ \varepsilon^k\leq\frac{b-a}{c}\\$$ Adding all these inequalities, we get $$\frac{\varepsilon-\varepsilon^{k+1}}{1-\varepsilon}\leq k\left(\frac{b-a}{c}\right)$$ or, after taking the limit of the left hand side as $k\to\infty$ $$\lim_{k\to\infty}\frac{1}{k}\cdot\frac{\varepsilon-\varepsilon^{k+1}}{1-\varepsilon}=0\leq\frac{b-a}{c}$$ and so $a\leq b$.
  3. $c<0$. Let $c'=-c>0$. Then $$a+c\varepsilon\leq b\implies a\leq b+c'\varepsilon\implies\frac{a-b}{c'}\leq\varepsilon.$$ By our hypothesis, this inequality holds for all $\varepsilon\in A:=(0,1)$. So $\frac{a-b}{c'}$ is a lower bound of $A$. But $\inf A=0$ and by definition of $\inf$, $\frac{a-b}{c'}\leq \inf A=0.$ It follows that $a\leq b$.
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  • $\begingroup$ Is $\epsilon$ arbitrary ? Or,$\epsilon$ arbitrary $\rightarrow \delta$ arbitrary ? I don't understand "$\delta$ is arbitrary". $\endgroup$
    – Leonardo Z
    Apr 14, 2023 at 8:21
  • $\begingroup$ @LeonardoZ, I have clarified the third case. Actually, now that I think about it, the case $c>0$ can be determined using the properties of $\inf$ and $\sup$, just like the third case. $\endgroup$
    – tmaj
    Apr 14, 2023 at 10:37
  • $\begingroup$ $\epsilon$ is not arbitrary, $\epsilon \in [0,1]$ specifically. $\endgroup$
    – Snared
    Apr 14, 2023 at 10:51

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