How to prove that $n^n \times n! \times (\sqrt{2})^{n}\le (2n)! $

Question

I would appreciate if somebody could help me with the following problem:

Q: How to prove that ($n \in \mathbb{N}$)
$$n^n \times n! \times (\sqrt{2})^{n}\le (2n)! $$

I try

$\log(n^n \times n! \times (\sqrt{2})^{n})\le \log(2n)!$

$\to$

$n\log n +\log n!+ n\log \sqrt{2} \le \log(2n)!$

$\to$

$n(\log n +\log \sqrt{2}) \le \log(n+1)+ \log(n+2)+\cdots+ \log (2n)$

And then?

Answer 1

Hint: Prove that for all $1 \leq k \leq n$ you have $$2n^2 \leq (n+k)(2n+1-k)$$

Answer 2

I would say your best bet is Proof by Induction.

First note the two sides are never equal unless you make the incorrect assumption that $0^0=1$.

Base case of $n=1$: $$LHS\to 1\cdot1\cdot(\sqrt2)^1=\sqrt2$$ $$RHS\to 2!=2>\sqrt2$$ so true for $n=1$

Inductive Step: Assume true for $n=k$: $$k^k\cdot k!\cdot(\sqrt2)^k<(2k)!$$ And use to show true for $n=k+1$

So prove using the above statement that: $$(k+1)^{k+1}\cdot(k+1)!\cdot(\sqrt2)^{k+1}<(2k+2)!$$ It may help to re-write it as: $$(k+1)(k+1)^{k}+(k+1)(k!)+\sqrt2(\sqrt2)^k<(2k+2)(2k+1)(2k)!$$

Answer 3

Induction will do the trick if you're willing to prove a bunch of base cases (specifically, as we'll see, for $n=1$ to $13$) and if you know that $(1+{1\over n})^n\lt e$ for all $n$ and $\sqrt2e\approx3.844231\lt4-{1\over7}$. The crucial induction step is

$$\begin{align} (n+1)^{n+1}(n+1)!(\sqrt2)^{n+1} &=\sqrt2\left(1+{1\over n}\right)^n(n+1)^2n^nn!(\sqrt2)^n\\ &=\sqrt2\left(1+{1\over n}\right)^n(n+1)^2(2n)!\quad\text{by the induction hypothesis}\\ &\lt\left(4-{1\over7}\right)(n+1)^2(2n)!\\ &\lt\left(4-{2\over n+1}\right)(n+1)^2(2n)!\quad\text{if }n\ge13\\ &=(4n+2)(n+1)(2n)!\\ &=2(2n+1)(n+1)(2n)!\\ &=(2n+2)(2n+1)(2n)!\\ &=(2n+2)! \end{align}$$

There might be some nice way to avoid the plethora of base cases here -- I'm not thrilled with the idea of having to verify the inequality all the way up to $n=13$ (and, admittedly, have not done so myself) -- but I don't see any obvious alternative. Maybe someone else does.