Show that $2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$ Is there a way to show that (with integer $n&gt;2$) $$2^{\lceil n \log_23\rceil}-3^n&lt;3^n-2^n$$
I tried to figure a way with derivative or by looking how both side are growing, but I have some trouble with the ceiling function.
With $n+k= {\lceil n \log_23\rceil}&lt;n \log_23+1$ we can use the Rhin bound (page 160):
$$|(n+k) \log2 - n \log3|&gt;\frac 1{(n+k)^{13.3}}&gt;\frac1{(n \log_23+1)^{13.3}}&gt;\frac 1{n^a}$$
If we choose $a=15$ the above is true for $n&gt;41$
Now, $$\frac 1{n^a}&gt;\frac{1}{(\frac{3}{2})^n}$$
is true for $n&gt;\frac{-aW_{-1}(\frac{-\log\frac{3}{2}}{a})}{log(\frac{3}{2})}$ or $n&gt;196$ with chosen $a=15$ ($W$ is the productlog), and we also have
$$\frac{1}{(\frac{3}{2})^n}&gt;\log(1+\frac{1}{(\frac{3}{2})^n})=\log(1+\frac{2^n}{3^n})$$
So for $n&gt;196$ we have $$|(n+k)\log2 - n \log3|=\log(\frac{2^{n+k}}{3^n})&gt;\log(1+\frac{2^n}{3^n})$$
or
$$\frac{2^{n+k}}{3^n}&gt;1+\frac{2^n}{3^n}$$
$$2^{\lceil n \log_23\rceil}-3^n&gt;2^n$$
Similarly we have
$$|n \log3-(n+k-1) \log2|&gt;\frac 1{(n+k-1)^{13.3}}&gt;\frac1{(n \log_23)^{13.3}}&gt;\frac 1{n^a}$$
which also holds for $a=15$, and additionally we have:
$$\frac{1}{(\frac{3}{2})^n}=\frac{1}{2^{n\log_2\frac{3}{2}}}&gt;\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}&gt;\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
So for $n&gt;196$ we have
$$|n \log3-(n+k-1) \log2|=\log(\frac{2\cdot 3^n}{2^{n+k}})&gt;\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
or
$$\frac{2\cdot 3^n}{2^{n+k}}&gt;1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}$$
$$2\cdot 3^n&gt;2^{n+k}+2^n$$
$$2^{\lceil n \log_23\rceil}-3^n&lt;3^n-2^n$$
This leads to ($n&gt;196$):
$$\begin{array}{|c|}\hline 2^n&lt;2^{\lceil n \log_23\rceil}-3^n&lt;3^n-2^n\\\hline\end{array}$$
and with manual checking, the left inequality holds except for $n=\{1,3,5\}$ and the right inequality holds except for $n=\{2\}$
Start by showing that
$$2^{\lceil \log_23^n\rceil}=2^{\lceil n \log_23\rceil}&lt;3^n$$
Then the rest follows!