If $n$ be a positive integer $>1$, prove that $$2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$$

Please help me to prove the above. I have to use laws of inequality like AM-GM. But how to use it for this particular problem.


Only use laws of inequality.

Edit 2

I want to solve this by using laws of inequality like weighted AM-GM. My attempt is the following

Consider positive numbers $\left(\frac{n}{1}\right), \left(\frac{n-1}{2}\right), \left(\frac{n-2}{3}\right), \cdots \left(\frac{2}{n-1}\right), \frac{1}{n}$ with corresponding weights $n, n-1, n-2, \cdots 2,1$, respectively and applying weighted AM>GM, we get,

$$\frac{\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}}{n+(n-1)+\cdots +2+1}>\left[\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}\right]^{\frac{n(n+1)}{2}}$$

I am unable to get the result because I am unable to get the sum $\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}$

Please suggest me some possible approach.

  • $\begingroup$ I think I proved it by induction, have you tried that? $\endgroup$ – Gregory Grant Feb 14 '16 at 5:49
  • $\begingroup$ @GregoryGrant It is a problem of Inequalities chapter. Please suggest some way by using the laws of inequality. $\endgroup$ – user1942348 Feb 14 '16 at 6:03
  • $\begingroup$ You can use laws of inequalities in induction. $\endgroup$ – fleablood Feb 14 '16 at 6:54
  • $\begingroup$ Nice inequality. (+1) $\endgroup$ – hypergeometric Feb 15 '16 at 15:23
  • $\begingroup$ My second proof is by induction. $\endgroup$ – marty cohen Feb 15 '16 at 17:54

Notice that $$ RHS = (n+1)^{n+1} \cdot \left(\frac{n}{1}\right) \cdot \left(\frac{n}{1}\cdot \frac{n-1}{2}\right) \cdots \left(\frac{n}{1}\cdot \frac{n-1}{2} \cdots \frac{1}{n}\right) = \prod_{k=1}^n \binom{n}{k} = \prod_{k=0}^n \binom{n}{k}. $$ Then, by AM-GM, $$ RHS = (n+1)^{n+1} \cdot \prod_{k=0}^n \binom{n}{k} \le (n+1)^{n+1} \cdot \left( \frac{\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{n}}{n+1}\right)^{n+1} = LHS. $$ Equality holds only if $\binom{n}0=\ldots=\binom{n}{m}$; i.e. for $n=1$.


Here is a proof by induction that starts with my formula for the right side.

The right side is $s(n) =(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1} $.

$\begin{array}\\ s(n) &=(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}(n-k+1)^{n-k+1}}{\prod_{k=1}^{n} k^{n-k+1}}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^k}{\prod_{k=1}^{n} k^{n-k+1}}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{\prod_{k=1}^{n} k^{n+1}}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{(n!)^{n+1}}\\ \end{array} $


$\begin{array}\\ \frac{s(n+1)}{s(n)} &=\frac{(n+2)^{n+2}\frac{\prod_{k=1}^{n+1}k^{2k}}{((n+1)!)^{n+2}}}{(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{(n!)^{n+1}}}\\ &=\frac{(n!)^{n+1}(n+2)^{n+2}(n+1)^{2(n+1)}}{((n+1)!)^{n+2}(n+1)^{n+1}}\\ &=\frac{(n!)^{n+1}(n+2)^{n+2}(n+1)^{n+1}}{((n+1)!)^{n+1}(n+1)!}\\ &=\frac{(n+2)^{n+2}(n+1)^{n+1}}{(n+1)^{n+1}(n+1)!}\\ &=\frac{(n+2)^{n+2}}{(n+1)!}\\ \end{array} $

Since, if $t(n) =2^{n(n+1)} $, $\frac{t(n+1)}{t(n)} =2^{(n+2)(n+3)-(n+1)(n+2)} =2^{2(n+2)} $, if we can show that $\frac{(n+2)^{n+2}}{(n+1)!} < 2^{2(n+2)} $, we are done.

This is $n+2 < 4((n+1)!)^{1/(n+2)} $ or, replacing $n+1$ by $n$, $n+1 < 4(n!)^{1/(n+1)} $.

Since $n! \gt (n/e)^n $, this is implied by

$\begin{array}\\ n+1 < 4(n/e)^{n/(n+1)}\\ or\\ (n+1)^{n+1} < 4^{n+1}(n/e)^n\\ or\\ (n+1)e^n(1+1/n)^n <4^{n+1}\\ or\\ (n+1)e^{n+1} <4^{n+1}\\ or\\ n+1 < (4/e)^{n+1}\\ or, \text{again replacing }n+1 \text{ by }n,\\ n < (4/e)^n\\ or\\ n^{1/n} < 4/e\\ \end{array} $

Since $n^{1/n} \le e^{1/e} < 1.45 $ and $4/e > 1.47 $, this is true for all $n$.

Therefore $\frac{s(n+1)}{s(n)} < \frac{t(n+1)}{t(n)} $ or $s(n+1) < t(n+1)\frac{s(n)}{t(n)} $.

Therefore, if $s(n) < t(n)$, $s(n+1) < t(n+1)$.

Since $s(2) =3^32^2\frac12 =54 $ and $t(2) =2^{6} =64 \gt s(2) $, $t(n) > s(n)$ for $n \ge 2$.


I will show that

$2^{n(n+1)} \gt (n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n} $

for $n \ge 16$.

The right side is $s(n) =(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1} $.

$\begin{array}\\ s(n) &=(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}(n-k+1)^{n-k+1}}{\prod_{k=1}^{n} k^{n-k+1}}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^k}{\prod_{k=1}^{n} k^{n-k+1}}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{\prod_{k=1}^{n} k^{n+1}}\\ &=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{(n!)^{n+1}}\\ \end{array} $

so, if $c=\frac12\ln(2\pi) \approx 0.92 $,

$\begin{array}\\ \ln(s(n)) &=(n+1)(\ln(n+1)-\ln(n!))+2\sum_{k=1}^{n}k\ln(k)\\ &\sim (n+1)(\ln(n+1)-\ln(n!))+2(\frac12 n^2\ln(n)-n^2/4+n\,\ln(n)/2)\\ &\sim (n+1)(\ln(n)+\frac1{n}-c-\frac12\ln(n)-n\ln(n)+n)+ n^2\ln(n)-n^2/2+n\,\ln(n)\\ &= \ln(n)(n+1-c-\frac12-n(n+1)+n^2+n) +1+n(n+1)-n^2/2\\ &= \ln(n)(n+\frac12-c) +(n^2+2n+2)/2\\ &= (n-.42)\ln(n) +(n^2+2n+2)/2\\ \end{array} $

If $t(n) =2^{n(n+1)} $, $\ln(t(n)) =n(n+1)\ln(2) $, so we are done if $(n-.42)\ln(n)+(n^2+2n+2)/2 < n(n+1)\ln(2) $. According to Wolfy, this is true for $n \ge 16$.

  • $\begingroup$ Thou shalt never mix $\sim$ with explicit inequalities. It might turn out to be true at $n=16$, but this argument doesn't establish that. $\endgroup$ – Greg Martin Feb 14 '16 at 7:40
  • $\begingroup$ @martycohen Nice and thanks for the effort. I am happy. Although I am still interested to use weighted AM-GM with suitable choice of terms. $\endgroup$ – user1942348 Feb 14 '16 at 8:17
  • $\begingroup$ I came up with a proof by induction and submitted it as another answer. $\endgroup$ – marty cohen Feb 14 '16 at 17:00

Take $\log$ of both sides and verify that the inequality becomes equality for $n=1$. Take backward difference:

$$\nabla RHS=(n+1)\log (n+1)- n\log n+n\log n-\sum_{i=1}^n \log i\le 2n\log 2 =\nabla LHS \tag 1$$ holds for $n=1$. Take another backward difference $$\nabla^2 RHS=(n+1)\log(1+\frac 1 n)\le 2\log 2 =\nabla^2 LHS\tag 2$$ which holds for $n=1$ as well. Since $((x+1)\log(1+\frac 1 x))'=\log(1+\frac 1 x)-\frac 1 x<0$ on $x>0$, $(2)$ is a strict inequality for $n>1$ hence $(1)$ is a strict inequality for $n>1$ and hence the original (strict) inequality holds for $n>1$. Q.E.D.

ETA: The above also shows that $\frac {LHS}{RHS}\to 2\log 2\approx 1.386$ and possibly monotonically.


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