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.?

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.
