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This is from the book How to think like a Mathematician,

How can I prove the inequality $$\sqrt[\large 7]{7!} < \sqrt[\large 8]{8!}$$

without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$

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  • $\begingroup$ Are we allowed complex roots and negative roots to prove the contrary ;-) $\endgroup$ Feb 28, 2015 at 19:22
  • $\begingroup$ I don't think that it is in the proposition $\endgroup$
    – Gwydyon
    Mar 2, 2015 at 11:24
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    $\begingroup$ 7! is the product of seven positive numbers, all of which are smaller than 8. $8^7$ is the product of 7 eights. So the inequality follows. $\endgroup$ Dec 7, 2021 at 22:52
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    $\begingroup$ Be sure to write your argument starting from $7! < 8^7$ and ending with what you want to prove, not backwards the way you discovered where to start. $\endgroup$ Dec 7, 2021 at 22:53
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    $\begingroup$ If you have a dataset of $n$ real numbers, and you add an $n+1$st number greater than the average of the $n$ numbers, the new set has a greater average. (Hint: logarithms.) $\endgroup$ Dec 7, 2021 at 22:53

7 Answers 7

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Your inequality is equivalent to $$(7!)^8 < (8!)^7$$ divide it by $(7!)^7$, and get $$7! < 8^7$$ and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$

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    $\begingroup$ In the spirit of the question title, I'd note that the first line of this answer is key. Maths does of course love cleverness, but also thrives on knowing to do really simply things such as taking powers of both sides here to remove those ugly roots. $\endgroup$
    – Keith
    Feb 27, 2015 at 3:02
  • $\begingroup$ You are right, great ! I would have of to think there. $\endgroup$
    – Gwydyon
    Feb 28, 2015 at 17:39
  • $\begingroup$ Elegant! really clever! $\endgroup$
    – Max Payne
    May 27, 2015 at 16:19
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Think of

$${\ln(7!)\over7}={\ln(1)+\cdots+\ln(7)\over7}$$

as the average of seven numbers and

$${\ln(8!)\over8}={\ln(1)+\cdots+\ln(8)\over8}$$

as the average when an eighth number is added. Since the new number is larger than the previous seven, the average must also be larger. (E.g., if you get a better score on your final than on any of your midterms, your grade should go up, not down.)

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  • $\begingroup$ You are right, but it is clear to me only when i compute ln(8) $\endgroup$
    – Gwydyon
    Feb 28, 2015 at 18:08
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    $\begingroup$ @Gwydyon, the logarithm is an increasing function, so $\ln(8)$ is larger than its predecessors. $\endgroup$ Feb 28, 2015 at 20:00
  • $\begingroup$ Yes I know that. $\endgroup$
    – Gwydyon
    Mar 2, 2015 at 11:21
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    $\begingroup$ @Gwydyon, I guess I don't understand your previous comment then. $\endgroup$ Mar 2, 2015 at 15:22
  • $\begingroup$ Because x/8 is lower than y/7, if x=y, that is not the case but ln(7!) and ln((7+1)!) are near close, for x>=1 $\endgroup$
    – Gwydyon
    Mar 3, 2015 at 16:41
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Note that $$ \sqrt[7]{7!} < \sqrt[8]{8!} \iff\\ (7!)^8 < (8!)^7 \iff\\ 7! < \frac{(8!)^7}{(7!)^7} \iff\\ 7! < 8^7 $$ You should find that the proof of this last line is fairly straightforward.

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    $\begingroup$ You are right but someone has writen the same statements $\endgroup$
    – Gwydyon
    Feb 28, 2015 at 18:11
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$8\ln (7!) < 7\ln (8!) \Rightarrow \ln (7!) < 7\ln 8 \iff \ln 1 + \ln 2 +\cdots \ln 7 < 7\ln 8$ which is clear.

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  • $\begingroup$ great! you achieved my second statement $\endgroup$
    – Gwydyon
    Feb 28, 2015 at 18:53
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You have already turned the comparison of two geometric means into the comparison of two arithmetic means. So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest. \begin{equation*} \begin{split} (1/(n+1)) \sum_{i=1}^{n+1} x_i &= (1/(n+1)) (x_{n+1} + \sum_{i=1}^{n} x_i) \\ &=(1/(n+1) (n x_{n+1}/n + n \sum_{i=1}^{n} x_i / n) \\ &> (1/(n+1) (\sum_{i=1}^{n} x_i/n + n \sum_{i=1}^{n} x_i / n) \\ &= (1/(n+1) ((n+1) \sum_{i=1}^{n} x_i / n) \\ &= \sum_{i=1}^{n} x_i / n \end{split} \end{equation*}

Note that we really only needed $x_{n+1}$ to be larger than the previous mean.

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The solution occurs just by doing simple calculations,
Lets start, $\sqrt[7]{7!}<\sqrt[8]{8!}$
iff $(\sqrt[7]{7!})^{7\cdot8}<(\sqrt[8]{8!})^{7\cdot8}$
iff $(7!)^{8}<(8!)^7$
iff $(7!)^8<(7!\cdot8)^7$
iff $(7!)^8<8^7\cdot(7!)^7$
iff $(7!)<8^7$
iff $1\cdot2\cdots6\cdot7<8\cdot8\cdot8\cdot8\cdot8\cdot8\cdot8$
wich is obviously true since $1<8,2<8,\ldots,7<8$

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Consider this

$$ x=\ln 8!-\frac87\ln7!=\ln8-\frac17\ln7!=\frac17\left(7\ln8-\ln7!\right)=\frac17\ln\frac{8^7}{7!}>0 $$ Hence, since the exponential is a monotonically increasing function: $e^{x/8}>1\implies (8!)^{1/8}>(7!)^{1/7}.$

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