# Without using a calculator, is $\sqrt{8!}$ or $\sqrt{9!}$ greater?

Which is greater between $$\sqrt{8!}$$ and $$\sqrt{9!}$$?

I want to know if my proof is correct...

\begin{align} \sqrt{8!} &< \sqrt{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?

• Some good general strategies for determining $a ≷ b$ are calculating the sign of $a - b$, or if you know they have the same sign, whether $\frac{a}{b}$ is greater than or less than 1.. May 1, 2018 at 22:46
• Comments nuked. A civil tone is required on this site. If you wish to continue this discussion, please do so in a group chat. May 3, 2018 at 15:50
• $\sqrt[n]{n!}$ is the geometric average of the integers from $1$ to $n$. This is obviously a growing function.
– user65203
May 3, 2018 at 19:59
• @YvesDaoust Your comment is the best answer here. May 25, 2018 at 3:14

$$(\sqrt{8!})^ {72}= (8!)^9 = (8!) (8!)^8$$

$$(\sqrt{9!})^ {72} = (9!)^8 = (9\times 8!)^8 = 9^8 (8!)^8$$

The second one, wins hands down.

• This relies on the fact that $$(9^8 > 8!)$$. Which is true, but perhaps not obvious? May 2, 2018 at 3:34
• @RossPresser Could you explain what you mean by not obvious? May 2, 2018 at 3:54
• Both products have eight factors, and all of the factors in one product are larger than all the factors in the other, clearly the larger product has the larger factors.
– Nij
May 2, 2018 at 4:21
• @RossPresser, One side is $9*9*9*9*9*9*9*9$, the other is $8*7*6*5*4*3*2*1$. The first one of these is clearly far larger May 2, 2018 at 10:38

You are implicitly writing that $$\frac{(8!)^{(9/72)}}{(9!)^{(8/72)}} = \left( \frac{8!}{9!} \right)^{(1/72)}$$ which is wrong.

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.

• Same idea as mine, so +1 :) May 1, 2018 at 19:16
• You can also say that $\sqrt{8!}$ is the geometric mean of the numbers $\{1,2,3,\ldots,8\}$ while $\sqrt{9!}$ is the geometric mean of $\{1,2,3,\ldots,9\}$. And you expect the this mean to increase when you throw in one more number which is greater than all the old ones. May 1, 2018 at 22:09

We have obviously $8!<9^8$. Hence, it follows that $(9!)^8=(8!\cdot9)^8=(8!)^8\cdot 9^8>(8!)^8\cdot(8!)=(8!)^9$. This implies that $(9!)^{1/9}>(8!)^{1/8}$.

• Upvoted this as the best example of starting from a known fact and deriving the theorem, rather than assuming what we are trying to prove, which was the first and most important mistake the OP made. May 1, 2018 at 22:41

This step doesn't look right to me: \begin{gather} (8!)^{(9/72)} - (9!)^{8/72} < 0 \\[6px] (9!)^{8/72} \left(\left( \frac{8!}{9!} \right)^{(1/72)} - 1\right) < 0 \end{gather} When you divide $(8!)^{9/72}$ by $(9!)^{8/72}$, you should get $$\frac{(8!)^{9/72}}{(9!)^{8/72}} = \frac{(8!)^{9/72}}{(9\cdot8!)^{8/72}} = \frac{(8!)^{1/72}}{(9)^{8/72}} = \frac{(8!)^{1/72}}{(9)^{1/9}} = \left( \frac{(8!)^{(1/8)}}{9} \right)^{1/9}$$

Also, note that your proof is basically of this form: \begin{gather} a < b \\[6px] a - b < 0 \\[6px] \left(\frac{a}{b} -1\right) < 0\\[6px] \frac{a}{b} < 1 \end{gather} You can skip several steps and just do \begin{gather} a < b \\[6px] \frac{a}{b} < 1 \end{gather}

$$(8!)^9=(8!)^8\cdot 8! < (8!)^8\cdot 9^8= (9!)^8$$

I would compare it by using logarithms:

$$\ln a = \frac{1}{8}(\ln1 + ...+\ln8)$$

$$\ln b = \frac{1}{9}(\ln1 + ...+\ln8+\ln9)=\frac{8}{9} \ln a + \frac{1}{9}\ln9$$

$$\ln b-\ln a=\frac{1}{9}(\ln9-\ln a)=\frac{1}{9} \ln\frac{9}{a}$$

$a$ is obviously less than 8 so:

$$\frac{9}{a} \gt 1$$

$$\ln\frac{9}{a}\gt0$$

$$\ln b - \ln a \gt 0$$

$$b \gt a$$

Instead of comparing the diference, comparing the ratio should be easy in such cases:

$$\left(\frac{8!^{1/8}}{9!^{1/9}}\right)^{8}=\frac{8!}{9!^{8/9}}=\frac{8!}{9!}(9!)^{1/9}=\left(\frac{9!}{9^9}\right)^{1/9}<1$$

this quantity is inferior than $1$ because $9!<9^9$

if $a$ is strictly positive real and $n$ is positive rational and also $a^n<1$ then $a<1$

which says:

$$\frac{8!^{1/8}}{9!^{1/9}}<1$$ QED.

For any growing sequence, the arithmetic mean of the first terms is a growing function.$^*$

By taking the logarithm, this extends to the geometric mean.

$$^*t_1,t_2,\cdots t_n<t_{n+1}\implies s_n<nt_{n+1}\implies\frac{s_n}n<\frac{s_n+t_{n+1}}{n+1}=\frac{s_{n+1}}{n+1}.$$

You need only note that $$9^8>8!,$$ then we'd be done. And $$9^8>8!$$ since all the eight factors on the left exceed all the eight factors on the right.

Then multiplying both sides by $$(8!)^8$$ gives $$(8!)^89^8>(8!)^88!,$$ or $$(9!)^8>(8!)^8,$$ and by taking $$1/72$$nd powers, we obtain the result $$(9!)^{1/9}>(8!)^{1/8}.$$

• Sorry if I'm missing something, but I think this isn't what the question is asking for at all. Aug 29, 2019 at 22:34
• $8\not=8!$ and $9\not=9!$, though ... re-read the question. Aug 30, 2019 at 1:45
• @NoahSchweber Ugh, that's right. Thanks for bringing my notice to it. I corrected my answer. Aug 30, 2019 at 7:17
• @YiFan Thank you. I adjusted my answer. Aug 30, 2019 at 7:18