Show that $\forall n \in\Bbb N: e < \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right)$ 
Show that:
  $$
\forall n \in\Bbb N: e < \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right)
$$

Till now i've only worked out a couple of proof sketches, and I don't have an idea how to proceed with them. 
Please note that this question has already been asked here. The answer there uses derivatives and integrals which i'm not allowed to use.
First sketch
Consider the sequence:
$$
x_n = \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right)
$$
One of the ways to show what's required is to show that:
$$
x_{n+1} \le x_n
$$
namely the sequence is monotonically decreasing. Now given $n\in\Bbb N$ we may calculate $x_1$:
$$
x_1 = \left(1+{1\over 1}\right)^1\left(1+{1\over 2\cdot 1}\right) = 3
$$
Consider the limit:
$$
\lim_{n\to\infty}x_n = \lim_{n\to\infty} \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right) = e
$$
Now based on the fact that the sequence tends to $e$ and it is monotonically decreasing and $x_1 = 3$, then it should follow that:
$$
\forall n\in\Bbb N: x_n \ge e
$$
Here comes the hard part, I could't prove that $x_n$ is monotonically decreasing. I've considered the fraction:
$$
{x_{n+1}\over x_n} = \left(1 - {1\over (n+1)^2}\right)^n \cdot \frac{2n(2n+3)(n+2)}{(2n+2)(2n+1)(n+1)}
$$
Not sure how to show it is less than $1$.
Second sketch
This sketch is based on the idea from my previous question. Namely it has been shown there that:
$$
e \le \frac{n+2}{(n+1)(n+1)!} + \sum_{k=0}^n {1\over k!}
$$
It looks like:
$$
e \le \frac{n+2}{(n+1)(n+1)!} + \sum_{k=0}^n {1\over k!} \le \left(1+{1\over n}\right)^n \left(1 + {1\over 2n}\right) \tag1
$$
In such case if we prove $(1)$ we are done. I've given it several tries but it gets ugly very soon.
The question is:

Is it possible to utilize any of those sketches to finish the proof? If not what would be the way to prove the inequality using anything before the definition of a derivative/Taylor series/intergrals

Thank you!
 A: The first sketch works.
We need to prove that $$\frac{x_n}{x_{n+1}}>1$$ or
$$\frac{\left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{2n}\right)}{\left(1+\frac{1}{n+1}\right)^{n+1}\left(1+\frac{1}{2n+2}\right)}>1$$ or
$$\frac{(n+1)^n(2n+1)(n+1)^{n+1}(2n+2)}{n^n(n+2)^{n+1}2n(2n+3)}>1$$ or
$$\frac{(n+1)^{2n+2}}{n^{n+1}(n+2)^{n+1}}>\frac{2n+3}{2n+1}$$ or$$\left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1}>\frac{2n+3}{2n+1},$$
which is true because
$$\left(\frac{n^2+2n+1}{n^2+2n}\right)^{n+1}=\left(1+\frac{1}{n^2+2n}\right)^{n+1}>1+\frac{n+1}{n^2+2n}+\frac{(n+1)n}{2(n^2+2n)^2}>\frac{2n+3}{2n+1}.$$
A: $$\frac{\log(1+x)}{x}+\log\left(1+\frac{x}{2}\right)=1+\frac{5x^2}{24}-\frac{5x^3}{24}+\frac{59 x^4}{320}-\ldots=1+\sum_{m\geq 2}a_m x^m $$
for any $x\in(0,1)$, where the sequence $\{a_m\}_{m\geq 2}$ can be proved to have alternating signs and the sequence $\{|a_m|\}_{m\geq 2}$ can be checked to be weakly decreasing by induction on $m$. It follows that the LHS is $>1$ for any $x\in(0,1)$: your inequality follows by exponentiating both sides (with $x$ being chosen as $\frac{1}{n}$) and by checking $n=1$ as a separate case.
A: Another short proof exists, if you use $~\displaystyle\left(1+\frac{1}{n}\right)^{\frac{1}{2}}<1+\frac{1}{2n}~$ and $~\displaystyle e<\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}$ 
which is proofed in the note of my question here .
