Find the smallest value of $\alpha\in \mathbb{R}$ such that for all $x>0$ you have $\left(1+\frac{1}{x}\right)^{x+\alpha}>e$ Find the smallest value of $\alpha\in \mathbb{R}$ such that for all $x>0$ you have
$$\left(1+\frac{1}{x}\right)^{x+\alpha}>e$$
By now i have tried the usual, taking logarithm and trying to solve for alpha, there i get the next condition:
$$\alpha>\frac{1-\ln\left(1+\frac{1}{x}\right)^x}{\ln\left(1+\frac{1}{x}\right)}$$
so the problem now is to maximize this function, but i didn't get it through standar ways. I know that $\lim_{x\to \infty}f(x)=\frac12$ (where $f$ is the last function), but to say that $\frac12$ is the supremum i need to prove that it's increasing or bounded by $\frac{1}{2}$ and i didn't got any of those two things.
 A: The smallest value is $1/2$.
Indeed, as @Gary outlined here, $\alpha=1/2$ works since \begin{align*}
\left( {x + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{x}} \right) &= (2x + 1)\tanh ^{ - 1} \left( {\frac{1}{{2x + 1}}} \right) \\ & = 1 + \frac{1}{{3(2x + 1)^2 }} + \frac{1}{{5(2x + 1)^4 }} +  \cdots  > 1.
\end{align*}
It remains to show that the inequality does not hold for $\alpha<1/2$. We have $$(x+\alpha)\log\left(1+\frac1x\right)>1\impliedby\log\left(1+\frac1x\right)-\frac1{x+\alpha}>0$$ and we find that the only critical point of the LHS occurs at $x^\ast=\alpha^2/(1-2\alpha)$ by setting the derivative to zero. At this value, we have \begin{align}\log\left(1+\frac1{x^\ast}\right)-\frac1{x^\ast+\alpha}&=\log\frac{(\alpha-1)^2}{\alpha^2}-\frac{1-2\alpha}{\alpha(1-\alpha)}\\&=2\log\left(\frac1\alpha-1\right)-\frac1\alpha+\frac1{1-\alpha}\\&=2\log\beta-\left(\beta-\frac1\beta\right)\end{align} where $\beta=1/\alpha-1$. It is now easy to show that this is positive only when $\beta<1$, or $\alpha>1/2$.
Note: a similar question was asked here: Find the smallest $a>0$ such that $(1+\frac{1}{x})^{x+a}>e$ for all $x\geq 1$ but it is not the same as we have to show the function is strictly decreasing over $(0,1)$.
