Let f(x) be a convex function st f(0)=0 and f(1)=1

Prove that if A > B, then we cannot have that for all c in (0,1);

A*f(c) + B*f(1-c) > B

I was actually given an answer to this, but can't seem to understand it, I was hoping for some help:

Here is the solution:

Proof by contradiction

The above statement is equivalent to:

$\displaystyle \frac{f(c)}{1-f(1-c)} > \frac{A}{B} $ for all c in (0,1)

But since f is convex, then (omitting a few steps):

$\displaystyle \frac{f(c)}{1-f(1-c)} < 1 $ for all c in (0,1)

Up to here I understand perfectly, the prof then seemed to make a giant leap by saying that since this implies that

$\displaystyle \frac{f(c)}{1-f(1-c)} $ can take on all values between in (0,1) for all c in (0,1), we can find a c* such that

$\displaystyle \frac{f(c*)}{1-f(1-c*)} < \frac{A}{B} $ and we have a contradiction

That's what I was given: no limits, continuity or derivatives.

Can someone see how that last step follows smoothly from what came before?