# Why does the following inequality hold?

Let $r$ be a positive integer greater than $1$ and set $$N_r = \dfrac{\log r +\log_2(r)}{\log 2}.$$

Why does it then follow that $$\displaystyle\dfrac{2^{n-1}}{n-1} \geq \log_2\left( \dfrac{r \log r}{\log r+\log_2 r}\right)$$
for $n \geq N_r + 1?$ The paper I am reading states that it follows because $\frac{2^t}{t}$ increases for $t \geq \frac{1}{\log 2}.$ But I do not see why it follows from this statement.

• Is there any typo? Is it $\frac{2^{n-1}}{n-1}\geq(\frac{r\log r}{\log r + \log_2 r})\log2$. – YHH May 15 '18 at 20:59
• @YHH thanks! You spotted the typo in the article. – inequalitynoob2 May 15 '18 at 21:07

Is there any typo? What I get is $\frac{2^{n-1}}{n-1}\geq(\frac{r\log r}{\log r + \log_2 r})\log2$.
Suppose the base for log is 10. Since $\frac{2^t}{t}$ is increasing for $t\geq 1 / \log 2$. It follows that $\frac{2^{n-1}}{n-1}\geq \frac{2^{N_r}}{N_r}=\frac{2^{\frac{\log r}{\log 2}} 2^\frac{\log_2 r}{\log 2}}{\log r + \log_2 r}\cdot \log 2$. Use change of base, $\frac{\log r}{\log 2}=\log_2r$. So $2^{\frac{\log r}{\log 2}}=r$. Thus, $\frac{2^{n-1}}{n-1}\geq \frac{2^{N_r}}{N_r}=\frac{r \cdot 2^\frac{\log_2 r}{\log 2}}{\log r + \log_2 r}\cdot \log 2 \geq \frac{r \cdot \log r}{\log r + \log_2 r}\cdot \log 2$.