If $\frac{a}{k} - \frac{a(c-1)}{kb} > 1$, is $\frac{a}{k+1} - \frac{a(c-1)}{(k+1)b} > 1$?

Question

I'm trying an induction method, but not sure if I'm able to prove it or find a counterexample.

Suppose $3 \leq a < b$ and $1 \leq c \leq b$, and $k \geq 1$, where $a, b, c, k$ are integers.

Suppose $n = 1$ is proven. Then inductive hypothesis is $$\frac{a}{k} - \frac{a(c-1)}{kb} > 1$$ Is the following true? $$\frac{a}{k+1} - \frac{a(c-1)}{(k+1)b} > 1$$

Answer 1

`Take $a-\frac{a(c-1)}{b}=k+\frac{1}{4}.$

We have $$\frac{a}{k}-\frac{a(c-1)}{kb}=1+\frac{1}{4k}>1,$$ but $$\frac{a}{k+1}-\frac{a(c-1)}{(k+1)b}=\frac{k+\frac{1}{4}}{k+1}<1.$$