# Proving $a \leq b$ for all $\epsilon \in (0,1)$: need help with inequality involving real numbers $a$, $b$, and $c$.

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.

• I think you are right.Thank you for your help. Apr 14, 2023 at 2:22

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.

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 $$a\leq b$$.
• Is $\epsilon$ arbitrary ? Or,$\epsilon$ arbitrary $\rightarrow \delta$ arbitrary ? I don't understand "$\delta$ is arbitrary". Apr 14, 2023 at 8:21
• @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.
– tmaj
Apr 14, 2023 at 10:37
• $\epsilon$ is not arbitrary, $\epsilon \in [0,1]$ specifically. Apr 14, 2023 at 10:51