Can We Prove That ##A<B(\epsilon-\delta)##?

[amirmath]

For arbitrary positive numbers $\epsilon$ and $\delta$ we know that $0<\delta<\epsilon$ such that $0<A<B(\epsilon-\delta)+\epsilon C$ for A, B, C>0. Can we conclude $A<B(\epsilon-\delta)$?

 

[Office_Shredder]

When you say arbitrary, do you mean it holds for every choice of epsilon and delta, or there exists some epsilon and delta? Because for the latter it's clearly not true, and for the former no such numbers A,B,C exist. For example if \delta = \epsilon/2 you get
0&lt;A&lt; \epsilon( B/2+C)
And sending epsilon to 0 shows that A has to be zero which is a contradiction to it being positive

 

[amirmath]

Thanks for your comment. I mean that the above tow inequalities hold for for every choice of epsilon and delta.

 

[Mark44]

Thanks for your comment. I mean that the above tow inequalities hold for for every choice of epsilon and delta.

Here at Physics Forums, surround your expressions with pairs of $$ or ##. A single $ doesn't do anything.

 

[amirmath]

Thank you Mark44