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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Are we allowed complex roots and negative roots to prove the contrary ;-) – Philip Oakley Feb 28 '15 at 19:22
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I don't think that it is in the proposition – Gwydyon Mar 02 '15 at 11:24
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Can these simpler approaches to a specific example be used to produce a simpler answer to the "general case"? – Mark Hurd Mar 03 '15 at 03:02
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I don't know, the problem was "as is" – Gwydyon Mar 03 '15 at 16:35
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More general inequality: Prove nth root of n! is less than n+1 th root of ((n+1) !): $\sqrt[n]{n!}\lt \sqrt[n+1]{(n+1)!}$? – Martin Sleziak Dec 11 '16 at 11:36
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127! 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. – Mark Saving Dec 07 '21 at 22:52
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3Be 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. – Ethan Bolker Dec 07 '21 at 22:53
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2If 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.) – Thomas Andrews Dec 07 '21 at 22:53
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$7!=5040, 8^7>8^5=8^48=2^{12}8=4048*8$. – markvs Dec 07 '21 at 23:03
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@markvs $2^{12}=4096$ (which only improves your argument). – Andreas Blass Dec 08 '21 at 00:23
7 Answers
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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 $$1 \cdots 7 < 8 \cdots 8$$
Crostul
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19In 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. – Keith Feb 27 '15 at 03:02
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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.)
Barry Cipra
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5@Gwydyon, the logarithm is an increasing function, so $\ln(8)$ is larger than its predecessors. – Barry Cipra Feb 28 '15 at 20:00
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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 – Gwydyon Mar 03 '15 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.
Ben Grossmann
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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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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 $1<8,2<8,\ldots,7<8$
Giorgos Kosmas
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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.
Alan
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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: $e^{x/8}>1\implies (8!)^{1/8}>(7!)^{1/7}.$
minmax
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