<s>[INST] 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 $n>2$) $$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$
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.
 [/INST] With $n+k= {\lceil n \log_23\rceil}<n \log_23+1$ we can use the Rhin bound (page 160: $|µ_1\log 2+µ_2\log 3|\geq H^{-13.3}$ with $H=max(|µ_1|,|µ_2|)$)
$$|(n+k) \log2 - n \log3|>\frac 1{(n+k)^{13.3}}>\frac1{(n \log_23+1)^{13.3}}>\frac 1{n^a}$$
If we choose $a=15$ the above is true for $n>41$
Now (growth of $x^a$ vs $a^x$), $$\frac 1{n^a}>\frac{1}{(\frac{3}{2})^n}$$
is true for $n>\frac{-aW_{-1}(\frac{-\log\frac{3}{2}}{a})}{log(\frac{3}{2})}$ or $n>196$ with chosen $a=15$ ($W$ is the productlog), and we also have (using $x>log(1+x)$)
$$\frac{1}{(\frac{3}{2})^n}>\log(1+\frac{1}{(\frac{3}{2})^n})=\log(1+\frac{2^n}{3^n})$$
So for $n>196$ we have $$|(n+k)\log2 - n \log3|=\log(\frac{2^{n+k}}{3^n})>\log(1+\frac{2^n}{3^n})$$
or
$$\frac{2^{n+k}}{3^n}>1+\frac{2^n}{3^n}$$
$$2^{\lceil n \log_23\rceil}-3^n>2^n$$
Similarly we have
$$|n \log3-(n+k-1) \log2|>\frac 1{(n+k-1)^{13.3}}>\frac1{(n \log_23)^{13.3}}>\frac 1{n^a}$$
which also holds for $a=15$, and additionally we have:
$$\frac{1}{(\frac{3}{2})^n}=\frac{1}{2^{n\log_2\frac{3}{2}}}>\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
So for $n>196$ we have
$$|n \log3-(n+k-1) \log2|=\log(\frac{2\cdot 3^n}{2^{n+k}})>\log(1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}})$$
or
$$\frac{2\cdot 3^n}{2^{n+k}}>1+\frac{1}{2^{\lceil n\log_2\frac{3}{2}\rceil}}$$
$$2\cdot 3^n>2^{n+k}+2^n$$
$$2^{\lceil n \log_23\rceil}-3^n<3^n-2^n$$
This leads to ($n>196$):
$$\begin{array}{|c|}\hline 2^n<2^{\lceil n \log_23\rceil}-3^n<3^n-2^n\\\hline\end{array}$$
and with manual checking, the left inequality holds except for $n$ in $\{1,3,5\}$ and the right inequality holds except for $n$ in $\{1,2\}$
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