Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$

Question

Why is $(n+1)^{n-1}(n+2)^n>3^n(n!)^2$ for $n>1$

I can use $$(n+1)^n>(2n)!!=n!2^n$$ but in the my case, the exponent is always decreased by $1$, for the moment I don't care about it, I apply the same for $n+2$

$(n+2)^{n+1}>(2n+2)!!=(n+1)!2^{n+1}$

gathering everything together,

$(n+1)^{n-1}(n+2)^n=\frac{(n+1)^n(n+2)^{n+1}}{(n+1)(n+2)}>\frac{(n+1)(n!)^22^{2n+1}}{(n+1)(n+2)}$

$\iff(n+1)^{n-1}(n+2)^n>(n!)^2\times\frac{2^{2n+1}}{(n+2)}$

but $\frac{2^{2n+1}}{(n+2)}>3^n$ is not true for $n=2$

can you suggest another approach ?

Answer 1

Hint

Induction for the step $n+1$:

$$(n+2)^{n}(n+3)^{n+1}=\frac{(n+3)^{n+1}}{(n+1)^{n-1}}(n+1)^{n-1}(n+2)^{n}>3^n(n!)^2\frac{(n+3)^{n+1}}{(n+1)^{n-1}}$$

We may expect

$$\frac{(n+3)^{n+1}}{(n+1)^{n-1}}>3(n+1)^2 \quad (1)$$

in order to finish the induction.

Backing to $(1)$ we have an equivalent expression:

$$\left(\frac{n+3}{n+1}\right)^{n+1}>3 \Leftrightarrow \left(1+\frac{2}{n+1}\right)^{n+1}>3$$

Using the Bernoulli inequality $(1+x)^m \ge1+mx$ for $x>-1$.

Taking $m=n+1$ and $x=\frac{2}{n+1}$ we get what we want.

Answer 2

EDIT: This answer is wrong, because I mixed up my left and right-hand sides right at the end. I think it is salvageable, but it'll be quite a bit of work.

I'll do it without induction.

Rearrange: we want $\left(\frac{(n+1)(n+2)}{3}\right)^{n-1} \frac{n+2}{3} > (n!)^2$

We'll show that this actually holds if we remove the $\frac{n+2}{3}$ term (which is always $\geq 1$ anyway).

The right-hand side of the modified desired inequality is is $$2^2 \times 3^2 \times \dots \times n^2$$ with $n-1$ terms. The left-hand side is $$\frac{(n+1)(n+2)}{3} \times \dots \times \frac{(n+1)(n+2)}{3}$$ with $n-1$ terms again. But $\frac{(n+1)(n+2)}{3}$ is bigger than or equal to the $i^2$ term whenever $i$ is less than or equal to $\sqrt{\frac{(n+1)(n+2)}{3}}$, and that's $\leq n$ whenever $n > 2$. So if $n>2$, for every $i \leq n$ we have each right-hand term less than its corresponding left-hand term.

So we only need to check $n=1$ and $n=2$, and they're very easy.

Answer 3

Let's see if something simple works.

You want $(n+1)^{n-1}(n+2)^n >3^n(n!)^2 $.

First, $(n+1)^{n-1}(n+2)^n > n^{2n-1} $.

Second, since $(n/e)^n < n! < (n/e)^{n+1} $ (easily proved by induction from $(1+1/n)^n < e < (1+1/n)^{n+1}$), $3^n (n!)^2 < 3^n(n/e)^{n+1} = (3n/e)^n(n/e) $.

Therefore, if $n^{2n-1} > (3n/e)^n(n/e) $, we are done.

This is the same as $(ne/3)^n > n^2/e $ or $ne/3 > (n^2/e)^{1/n} = (n^{1/n})^2/e^{1/n} $.

But $n^{1/n} < e^{1/e} <1.5 $ and $e^{1/n} > 1$ so $(n^{1/n})^2/e^{1/n} < 1.5^2 = 2.25 $.

Therefore $ne/3 > (n^{1/n})^2/e^{1/n} $ if $ne/3 > 2.25$ or $n > 3\cdot 2.25/ e \approx 2.48 $.

Smaller $n$ easily verified.

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Someone commented "How do you know that $e^{1/e} < 1.5$?". My answer was "Calculator."

Here is a calculator-free answer.

But $n^{1/n} < e^{1/e} < \sqrt{3} $ (since $2 <e < 3$) and $e^{1/n} > 1$ so $(n^{1/n})^2/e^{1/n} < (\sqrt{3})^2 = 3 $.

Therefore $ne/3 > (n^{1/n})^2/e^{1/n} $ if $ne/3 > 3$ or $n > 3\cdot 3/ e = 9/e \approx 3.31 $.

Smaller $n$ easily verified.

Answer 4

prove: $(n+1)^{n−1}(n+2)^n>3^n(n!)^2=3^n(n!n!)$ for $n>1$

$n=2:3^14^2=48>3^2(2)(2)=36$

assume: $(n+1)^{n−1}(n+2)^n>3^n(n!n!)$

need to arrive at: $(n+2)^{n}(n+3)^{n+1}>3^{n+1}(n+1)!(n+1)!$

for lhs need to multiply by: ${{(n+2)^{n}(n+3)^{n+1}}\over {(n+1)^{n−1}(n+2)^n}} ={{(n+3)^n(n+3)(n+1)}\over {(n+1)^n}}=({{n+3}\over {n+1}})^n(n+3)(n+1)$

where $({{n+3}\over {n+1}})^n$ is greater than $3$ for $n=3$ and increasing for all $n\in N$.

for rhs need to multiply by: $3(n+1)(n+1)$. therefore lhs > rhs and

$(n+1)^{n−1}(n+2)^n>3^n(n!)^2$ for $n>1$ by mathematical induction.