Inequality related with e number

Question

In a number theory paper I saw such inequality:

For any $t>0,$ $(1+\frac{1}{t})^{t+\frac{1}{2}} > e$.

How to prove it? It is obvious for $t \geq 1$ but how to prove for arbitrarily small $t$?

Answer 1

Note that for any $t>0$, \begin{align*} \left( {t + \frac{1}{2}} \right)\log \left( {1 + \frac{1}{t}} \right) &= (2t + 1)\tanh ^{ - 1} \left( {\frac{1}{{2t + 1}}} \right) \\ & = 1 + \frac{1}{{3(2t + 1)^2 }} + \frac{1}{{5(2t + 1)^4 }} + \cdots > 1. \end{align*} Taking the exponential of each side yields the required inequality.