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

• Are we allowed complex roots and negative roots to prove the contrary ;-) – Philip Oakley Feb 28 '15 at 19:22
• I don't think that it is in the proposition – Gwydyon Mar 2 '15 at 11:24
• Can these simpler approaches to a specific example be used to produce a simpler answer to the "general case"? – Mark Hurd Mar 3 '15 at 3:02
• I don't know, the problem was "as is" – Gwydyon Mar 3 '15 at 16:35
• – Martin Sleziak Dec 11 '16 at 11:36

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

• 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. – Keith Feb 27 '15 at 3:02
• You are right, great ! I would have of to think there. – Gwydyon Feb 28 '15 at 17:39
• Elegant! really clever! – Max Payne May 27 '15 at 16:19

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.)

• You are right, but it is clear to me only when i compute ln(8) – Gwydyon Feb 28 '15 at 18:08
• @Gwydyon, the logarithm is an increasing function, so $\ln(8)$ is larger than its predecessors. – Barry Cipra Feb 28 '15 at 20:00
• Yes I know that. – Gwydyon Mar 2 '15 at 11:21
• @Gwydyon, I guess I don't understand your previous comment then. – Barry Cipra Mar 2 '15 at 15:22
• 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 3 '15 at 16:41

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.

• You are right but someone has writen the same statements – Gwydyon Feb 28 '15 at 18:11

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

• great! you achieved my second statement – Gwydyon Feb 28 '15 at 18:53

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.