# An inequality with exponents, factorials and nth roots!

Problem: Prove for natural numbers $$n > 2$$, $$(\sqrt{2!}-1)((3!)^{\frac{1}{3}}-\sqrt{2!})\cdots(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}) < \frac{n!}{(n+1)^n}.$$

My attempts: By AM-GM we get,

$$(\sqrt{2!}-1)((3!)^{\frac{1}{3}}-\sqrt{2!})\cdots(((n+1)!)^{\frac{1}{n+1}}-(n!)^{\frac{1}{n}}) < \left( \frac{((n+1)!)^{\frac{1}{n+1}}-1}{n}\right)^{n}$$.

So, now its enough to show, $$\left( \frac{((n+1)!)^{\frac{1}{n+1}}-1}{n}\right)^{n} < \frac{n!}{(n+1)^n}$$ But I don't know how to do that.

I also tried with induction, but it did not work

Jack D'Aurizio's answer doesn't have any explicit bound for what $$n$$ is large enough to cut off the case work. Here's a possible way around this.

EDIT:
I managed to prove this by elementary means. Below is my old approach using properties of the $$\Gamma$$ function.

I'll start with the form $$(n+1)((n+1)!)^{1/(n+1)}-n(n!)^{1/n}.

Using AG, it's simple to see that $$n!^{1/n}<(n+1)/2$$, so we can deal with the following inequalities instead: $$(n+1)((n+1)!)^{1/(n+1)}-n(n!)^{1/n}<2(n!)^{1/n}\\ (n+1)((n+1)!)^{1/(n+1)}<(n+2)(n!)^{1/n}\\ (n+1)^{n+1}(n+1)n!<(n+2)^{n+1}n!(n!)^{1/n}\\ n+1<\left(1+\frac{1}{n+1}\right)^{n+1}(n!)^{1/n}\\ n+1<\left(1+\frac{1}{n}\right)^n(n!)^{1/n}$$ In the last line I used the well-known inequality $$\left(1+\frac{1}{n+1}\right)^{n+1}\geq\left(1+\frac{1}{n}\right)^n$$. So now we have: $$\left(\frac{n^n}{(n+1)^{n-1}}\right)^n We can use induction to get rid of $$n!$$. The inequality cleary holds for $$n=2$$. Then: $$(n+1)!=(n+1)n! >(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n$$ We could complete the induction step if we knew that $$(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n\geq\left(\frac{(n+1)^{n+1}}{(n+2)^n}\right)^{n+1}\\ 1\geq\left(\frac{n+1}{n+2}\frac{(n+1)^{2n}}{n^n(n+2)^n}\right)^n\\ 1\geq\frac{n+1}{n+2}\frac{(n+1)^{2n}}{n^n(n+2)^n}$$ But rearranging a bit, we find: $$1\geq\left(\frac{n+1}{n+2}\right)^{n+1}\left(\frac{n+1}{n}\right)^n\\ \left(1+\frac{1}{n+1}\right)^{n+1}\geq\left(1+\frac{1}{n}\right)^n$$ This is the well-known inequality from before, so we may now conclude our induction step: $$(n+1)!>(n+1)\left(\frac{n^n}{(n+1)^{n-1}}\right)^n\geq\left(\frac{(n+1)^{n+1}}{(n+2)^n}\right)^{n+1}$$ And we are done.

OLD:
Define $$f(x)=x(x!)^{1/x}$$. The statement above amounts to proving $$f(x+1)-f(x) for $$x$$ large enough and checking by hand for all smaller $$n$$. Now suppose we managed to show that beyond a certain $$x$$, we had $$f'(x). Then by Lagrange's theorem there would be some $$\xi\in(x,x+1)$$ such that $$f(x+1)-f(x)=f'(\xi)<\xi To prove this, we'll have to make some pretty ugly bounds for $$f'$$. First: $$f'(x)=(x!)^{1/x}\left(1-\frac{1}{x}\log(x!)+\psi(x+1)\right)$$ Here $$\psi$$ is the digamma function. Wikipedia has the following estimates for $$x!$$: $$\sqrt{2\pi}x^{x+1/2}e^{-x}\leq x!\leq e^{1-x}x^{x+1/2}$$ Also, the digamma function page has the following inequality: $$\psi(x)<\log x-\frac{1}{2x}$$ Putting all this together for $$x>1$$: \begin{align} f'(x)&<(e^{1-x}x^{x+1/2})^{1/x}\left(1+\log (x+1)-\frac{1}{2(x+1)}-\frac{1}{x}\left(\frac{1}{2}\log(2\pi)+\left(x+\frac{1}{2}\right)\log(x)-x\right)\right)\\ & Here I dropped a bunch of negative terms in the second line. I also used $$\log(1+1/x)<1/x$$ in the third line. Define $$g(x)=e^{1/x-1}x^{1/2x}(2+1/x)$$. Then we want to prove $$g(x)<1$$. Apparently (I calculated this with Mathematica because I don't feel like differentiating $$g$$): $$g'(x)=-\frac{1}{2}e^{-1+1/x}x^{-3+1/2x}(1+4x+(2x+1)\log(x))$$ But this is negative for any $$x>1$$. Now we might numerically calculate that $$g(9)\approx0.98$$ and because $$g$$ is decreasing, we have $$g<1$$ for $$x\geq 9$$. Now all that remains is checking for all $$n<9$$.

I don't like this approach because it's too numerical at the end. At the same time, $$f(x+1)-f(x)$$ seems like a very well-behaved function - indeed, I'm sure there exists some way of going about this more analytically.

• B E A U T I F U L... well actually rather ugly but an actually rigorous answer! I wish that the digamma function and your g(x) were not involved however because the book that I pulled it from hasn't defined either of these... it's from a chapter which covers the AGM inequality and Bernoulli inequality. The text is "More Calculus of a Single Variable". Well.... I think that for now... this is the best we have. If I can derive a more clean proof I'll post it..... until then, I believe your solution is the ONLY real solution (after one checks n=1,...9 which we leave to the reader LOL). Jun 5, 2019 at 21:54
• Some thoughts... it might be helpful to split it up into cases. If x(n) equals the product on the LHS, then a good strategy might be the following: Suppose that for all $n\le N$ we have $x(n)<\frac{n!}{(n+1)^n}$. Then we could try proving the result by contradiction. It may help to split it up into cases. Case 1: Assume $x(N+1) = \frac{(N+1)!}{(N+2)^{N+1}}$ and derive a contradiction. Case 2: Repeat with $=$ replaced by $>$. Sometimes splitting this up helps. unfortunately, it hasn't helped me yet. Jun 5, 2019 at 21:58
• @Squirtle Also, which book is the one you got this from?
– J_P
Jun 6, 2019 at 1:28
• More calculus of a single variable Jun 6, 2019 at 1:35
• Oh, that's the title, I thought it was the chapter...
– J_P
Jun 6, 2019 at 10:52

The last inequality can be written as $$(n+1)\cdot\left((n+1)!^{\frac{1}{n+1}}-1\right)<n\cdot n!^\frac{1}{n}$$ or: $$(n+1)\cdot (n+1)!^{\frac{1}{n+1}} -n\cdot n!^\frac{1}{n}< n+1$$ that is straightforward to prove through Stirling's approximation: $n!\approx\frac{n^n}{e^n}\sqrt{2\pi n}$ gives that the LHS behaves like $\frac{2}{e}\cdot(n+1)$ and $\frac{2}{e}<1$.

• I recently asked this question as well... and this answer doesn't at all seem to obviously work. If I did something similar and the result failed on the order of 10^-11 for all n greater than 5. This is to say, I feel your answer is too cavalier.... and if it can be made more rigorous would you please do this. I spent several hours yesterday working on this, so it is not as if I haven't tried this approach or seriously considered it. Can you clear it up more, please? Jun 4, 2019 at 18:35
• This answer could be completed if this issue gets resolved
– user679638
Jun 5, 2019 at 10:08