Show $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$. I need to show that $e < \Big(1 + \frac 2 {2x+1}\Big)^{x+1}$ for all $x \ge 1$. 
This happens if $(x+1)\ln\Big(1 + \frac 2 {2x+1}\Big) > 1$ so let's study the function $f(x) = (x+1)\ln\Big(1 + \frac 2 {2x+1}\Big)$
Edit
We have abandoned the previous method of only using a characterization of $e$ with inequalities since we thought it was impractical to do it. Let's return instead to analysis.


*

*$f(1) = 2 \ln(\frac 5 3) > 1$

*$f'(x) = \ln(1+ \frac 2 {2x+1}) - \frac {4(x+1)}{(4x^2+8x+3)} < 0$ on $[1,+\infty[$

*$\lim f(x) = 1$
However how do I show formally 2.?
 A: Consider the function $f(y):=(1+\frac{2}{y})^{y+1}$. Clearly, $\lim_{y\rightarrow\infty}f(y)=e^2$. Moreover, $f$ is monotone decreasing since
\begin{align*}
f'(y)<0 &\iff \left({1+\frac{2}{y}}\right)\log\left({1+\frac{2}{y}}\right) < \left({\frac{2}{y}}\right)+\frac{1}{2}\left({\frac{2}{y}}\right)^2 \\
& \iff (1+t)\log(1+t) < t+\frac{1}{2}t^2 \quad \text{for }t=\frac{2}{y},
\end{align*}
where the last inequality can be verified for all $t> 0$ since equality holds for $t=0$ and the right-hand side grows faster than the left-hand side (just look at the derivatives). Choosing $y=2x+1$, we can thus deduce
\begin{align*}
e^2<\Big(1+\frac{2}{2x+1}\Big)^{2x+2}.
\end{align*}
Taking the square root, we conclude 
\begin{align*}
e<\Big(1+\frac{2}{2x+1}\Big)^{x+1}.
\end{align*}
Edit: Modified due to the oversight pointed out by GFauxPas.
A: Hint:
$$ \Big(1 + \frac 2 {2x+1}\Big)^{x+1}= \Big(1 + \frac{1+\frac{1}{2x+1}}{x+1}\Big)^{x+1}>\Big(1 + \frac{1}{x+1}\Big)^{x+1}$$
A: From $f(x) = (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)$ we compute the derivative.  
$$
\begin{eqnarray}
f'(x) = & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)' \\
= & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(\frac {2x+3} {2x+1}\Big)' \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{2(2x+1) - 2(2x+3)}{(2x+1)^2} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{-4}{(2x+1)^2} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {1} {2x+3} \Big( \frac{-4}{(2x+1)} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) - \frac {4(x+1)} {4x^2+8x+3}
\end{eqnarray}
$$
A: $\quad$The question currently asks for a formal demonstration that 
$$\ln\left(1+ \dfrac 2 {2x+1}\right) < \dfrac {4(x+1)}{(4x^2+8x+3)}  \text{ for  } x\in[1,+\infty).$$
$\quad$To show it's true for $x=1$, we must show $\ln\left(1+\dfrac23\right)<\dfrac{4(2)}{4+8+3}=\dfrac8{15}.$
By exponentiating both sides, this is equivalent to $\dfrac53<\exp\left(\dfrac8{15}\right).$ 
Well, $\dfrac53=\dfrac{375}{225}<\dfrac{377}{225}=1+\dfrac8{15}+\dfrac12\left(\dfrac{8}{15}\right)^2<\exp\left(\dfrac8{15}\right).$
$\quad$Furthermore, it can be shown that 
$$\lim_{x\to\infty}\ln\left(1+ \dfrac 2 {2x+1}\right) = \lim_{x\to\infty}\dfrac {4(x+1)}{(4x^2+8x+3)} =0.$$
$\quad$Finally, it can be shown that $\ln\left(1+ \dfrac 2 {2x+1}\right)$ and $\dfrac {4(x+1)}{(4x^2+8x+3)}$ are monotonically decreasing in the relevant interval.  If $0\le x _0<x_1,$ then $2x_0+1<2x_1+1,$ so $\dfrac1{2x_1+1}<\dfrac1{2x_0+1},$
so  $\ln\left(1+ \dfrac 2 {2x_1+1}\right) <  \ln\left(1+ \dfrac 2 {2x_0+1}\right),$ 
and $4x_1x_0^2+4x_0^2+5x_0<4x_0x_1^2+4x_1^2+5x_1; $ 
adding $8x_0x_1+3x_1+3x_0+3$ to both sides,
$4x_1x_0^2+8x_0x_1+3x_1+4x_0^2+8x_0+3<4x_0x_1^2+8x_0x_1+3x_0+4x_1^2+8x_1+3$
i.e., $4(x_1+1)(4x_0^2+8x_0+3)<4(x_0+1)(4x_1^2+8x_1+3),$ so $$\dfrac{4(x_1+1)}{(4x_1^2+8x_1+3)}<\dfrac{4(x_0+1)}{(4x_0^2+8x_0+3)}.$$
