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.?

From $f(x) = (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)$ we compute the derivative.  
$$
\begin{eqnarray}
f'(x) = & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(1 + \frac 2 {2x+1}\Big)' \\
= & (x+1)'\log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\log\Big(\frac {2x+3} {2x+1}\Big)' \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{2(2x+1) - 2(2x+3)}{(2x+1)^2} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {2x+1} {2x+3} \Big( \frac{-4}{(2x+1)^2} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) + (x+1)\frac {1} {2x+3} \Big( \frac{-4}{(2x+1)} \Big) \\
= & \log\Big(1 + \frac 2 {2x+1}\Big) - \frac {4(x+1)} {4x^2+8x+3}
\end{eqnarray}
$$
