Show $n \cdot \log_b r + \log_b \frac{r}{r-1} \le \lceil n \cdot \log_b r \rceil$

Question

Let $n$ be a natural number and $b, r > 1$ be two natural numbers, then I guess we have $$ n \cdot \log_b r + \log_b \frac{r}{r-1} \le \lceil n \cdot \log_b r \rceil. $$ where $\lceil x \rceil = \min\{ k \in \mathbb Z : x \le k \}$ is the smallest integer greater or equal to $x \in \mathbb R$. Any ideas how to prove this relation?

Answer 1

Not true. Take $n=4,b=10,r=3$. Then, we have $$\text{LHS}=4\log_{10}3+\log_{10}\frac 32=\log_{10}\frac{243}{2}\gt \log_{10}100=2=\lceil 4\log_{10}3\rceil=\text{RHS}$$

Answer 2

Also, $$n\log{_b} \ {r} + \log{_b} \ \frac{r}{r-1} \le \lceil n\log{_b} \ {r}\rceil + \log{_b} \ \frac{r}{r-1} \le \lceil n\log{_b} \ {r}\rceil + \log{_b} \ 2$$