Inequality from Chapter 5 of the book *How to Think Like a Mathematician*

Question

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 $$\frac 17 \cdot \ln(7!) < \frac 18 \cdot \ln(8!)$$

Answer 1

$8\ln (7!) < 7\ln (8!) \Rightarrow \ln (7!) < 7\ln 8 \iff \ln 1 + \ln 2 +\cdots \ln 7 < 7\ln 8$ which is clear.

Answer 2

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

Answer 3

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.

Answer 4

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.)

Answer 5

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.

Answer 6

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: $e^{x/8}>1\implies (8!)^{1/8}>(7!)^{1/7}.$

Answer 7

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 $1<8,2<8,\ldots,7<8$