<s>[INST] Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and  $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
 \sqrt[8]{8!} &< \sqrt[9]{9!} \\
 (8!)^{(1/8)} &< (9!)^{(1/9)} \\
 (8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
 (8!)^{(9/72)} - (9!)^{8/72} &< 0 \\
 (9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) &< 0 \\
 \left(\frac{8!}{9!}\right)^{(1/72)} - 1 &< 0 \\
 \left(\frac{8!}{9!}\right)^{(1/72)} &< 1 \\
 \left(\left(\frac{8!}{9!}\right)^{(1/72)}\right)^{72} &< 1^{72} \\
 \frac{8!}{9!} < 1 \\
 \frac{1}{9} < 1 \\
\end{align}
if it is not correct how it would be?
 [/INST] Another way to see this is to convert both sides into two "averages".


*

*"Uniform distribution on $\{\log1,\dots,\log8\}$": $$\frac{\log1 + \dots + \log8}{8}$$

*"Uniform distribution on $\{\log1,\dots,\log9\}$": $$\frac{\log1 + \dots + \log9}{9}$$


Intuitively, the later has a greater "expectation", so the result follows.  If you're not satisfied with this probabilistic interpretation, break the later into a sum of two terms and regroup them as
\begin{align}
& \frac{\log1 + \dots + \log8}{8} < \frac{\log1 + \dots + \log8}{9} + \frac{\log9}{9} \\
&\iff (\log1 + \dots + \log8) \left(\frac18-\frac19\right) < \frac{\log9}{9} \\
&\iff \frac{\log1 + \dots + \log8}{72} < \frac{\log9}{9} \\
&\iff \log1 + \dots + \log8 < 8 \log9
\end{align}
The last inequality is true since $\log$ is strictly increasing.
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