Inequality from Chapter 5 of the book *How to Think Like a Mathematician* 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 $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$
 A: $8\ln (7!) < 7\ln (8!) \Rightarrow \ln (7!) < 7\ln 8 \iff \ln 1 + \ln 2 +\cdots \ln 7 < 7\ln 8$ which is clear.
A: 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.)
A: 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.
A: 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 $1<8,2<8,\ldots,7<8$
A: 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.
A: Your inequality is equivalent to
$$(7!)^8 < (8!)^7$$
divide it by $(7!)^7$, and get
$$7! < 8^7$$
and this is clear, since $$1 \cdots 7 < 8 \cdots 8$$
A: 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: $e^{x/8}>1\implies (8!)^{1/8}>(7!)^{1/7}.$
