Question

Let a,b,c\in R, If for all \epsilon\in(0,1), a +c\products \epsilon\leq b, then a\leq b. How to prove?

Answer 1

Suppose a > b, then a -b >0. Suppose that c >0

Let
\[
\epsilon =\inf \{ \frac 1 2, \frac {a-b}{5 c}t\}\\
\epsilon \in (0,1)\\

\epsilon <\frac {a-b}c \\
c\,\epsilon <a -b\\
a +c\,\epsilon> a -c\,\epsilon >b\
\]
a contradiction.
Try to do it with c<0

Answer 2

For clarity, is it

Let \(a,b,c \in \mathbb{R}\). If for all \(\epsilon\in(0,1), \;\; a +c\cdot \epsilon\leq b\), then \(a\leq b\).
Or
Let \(a,b,c \in \mathbb{R}\). If for all \(\epsilon\in(0,1), \;\; (a +c)\cdot \epsilon\leq b\), then \(a\leq b\).

Answer 3

the t should not be in the inf in my post above.