1

Prove that $$ (n!)^2\le \left[\frac{(n+1)(n+2)}{6}\right]^n $$ holds for all $n\in\mathbb{Z^+}$.

I tried induction but there's no obvious way to go from $n$ to $n+1$.

Martin Sleziak
  • 53,687
John Smith
  • 1,261
  • What difficulty are you having with this exercise? –  Jan 21 '15 at 03:36
  • Tried induction? – jimjim Jan 21 '15 at 03:53
  • 1
    Yes, I tried induction but there's no obvious way to go from n to n+1. Please help – John Smith Jan 21 '15 at 04:32
  • The induction is tricky and IMHO the question could be reopened. Seeing I cannot now post an answer, here is a very brief hint. Assume the inequality for $n-1$; proving for $n$ comes down to showing$$6n^{n+1}\le(n+1)(n+2)^n\ .$$Expanding by the binomial theorem and taking just the first few terms gives$$RHS\ge6n^{n+1}+\frac13n^{n-1}(n-1)(n-8)$$which is greater than LHS for $n\ge8$. – David Jan 21 '15 at 04:36
  • 1
    @JohnSmith : voted to reopen, also add some of the work you have tried. – jimjim Jan 21 '15 at 04:39
  • Thanks guys, I reckon it will come out easily by brute force by expansion using the Binomial Theorem, although we would need to manually check the small cases. I'm not sure if there are any other ways to do it though, will Stirling's formula help? – John Smith Jan 21 '15 at 05:16
  • What is the stronger result you are claiming? – John Smith Jan 21 '15 at 05:54
  • Got it wrong way. Sorry. – André Nicolas Jan 21 '15 at 05:59
  • BTW it is good to give information what you have tried in your post and not in the comments. (That was probably cause of the downvotes and closure.) – Martin Sleziak Jan 21 '15 at 14:03
  • 1
    @JohnSmith Where did this problem come from by the way? I'm curious--I've never seen this inequality before. – Daniel W. Farlow Jan 22 '15 at 17:56
  • Here is another post mentioning this inequality: http://math.stackexchange.com/questions/600640/induction-nn1-n1n-and-n2-leq-left-fracn-12n-16 – Martin Sleziak Oct 18 '15 at 20:55

2 Answers2

2

Try to prove first that $$n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6.$$ (I have posted a separate question about this sum.)

Then use the AM-GM inequality for the sum on the LHS.

Community
  • 1
Martin Sleziak
  • 53,687
  • @user2345215 I suspected that AM-GM might help, but it took me a long time to get to this form. I was trying it first for $1^2+2^2+\dots+n^2$ and then for something like $1\cdot2+2\cdot3+(n-1)\cdot n+n$. The results were two week, so I though that I must distribute the summands in some way a bit more uniformly. – Martin Sleziak Jan 21 '15 at 14:33
  • Apparently, I was close to this, but I didn't realize it :( I knew the obvious result $(1\cdot n)(2\cdot (n-1))\ldots(n\cdot 1)\le (((n+1)/2)^2)^n$ and tried to improve it, but somehow the other AM-GM didn't occur to me, even though I was already using it $n$ times on the groups of 2 instead of one time on the $n$ groups. – user2345215 Jan 21 '15 at 14:36
  • @MartinSleziak Could you expound on how you would use the AM-GM inequality in this instance? – Daniel W. Farlow Jan 22 '15 at 18:20
  • @induktio AM-GM says that for any nonnegative real numbers $a_1,\dots,a_n$ we have $\sqrt[n]{a_1\cdots a_n} \le \frac{a_1+\dots+a_n}n.$ Just use it for $a_1=n\cdot1, a_2=(n-1)\cdot2, \dots, a_n=1\cdot n$. – Martin Sleziak Jan 22 '15 at 20:31
1

Here is an idea I had (by no means a complete answer but a start nonetheless), where I assume you have verified the base case and all that jazz: \begin{align} [(k+1)!]^2 &= (k+1)^2(k!)^2\\[1em] &\leq (k+1)^2\left[\frac{(k+1)(k+2)}{6}\right]^k\\[1em] &\leq\frac{(k+2)(k+3)}{(k+1)(k+1)}\cdot\left\{(k+1)^2\left[\frac{(k+1)(k+2)}{6}\right]^k\right\}\\[1em] &= \frac{(k+2)^{k+1}(k+3)(k+1)^k}{6^k}\\[1em] &\leq\frac{(k+2)^{k+1}(k+3)^{k+1}}{6^{k+1}}\qquad\left[\text{since}\quad\frac{(k+3)(k+1)^k}{6^k}\leq\frac{(k+3)^{k+1}}{6^{k+1}}\right]\\ &\vdots \end{align} where the inequality after the "since" is true for $k\geq 18$.

It's hardly optimal, but it is something at least. It introduces another inequality that may be approached via asymptotics or the like--the extra $(k+1)$ term really seems to get in the way of many lines of attack.

Good luck!

Daniel W. Farlow
  • 22,531