Using the Cauchy-Schwarz Inequality, we have $$ \begin{align} 1 &=\left(\int_n^{n+1}1\,\mathrm{d}x\right)^2\\ &\le\left(\int_n^{n+1}x\,\mathrm{d}x\right)\left(\int_n^{n+1}\frac1x\,\mathrm{d}x\right)\\ &=\left(n+\frac12\right)\log\left(1+\frac1n\right) \end{align} $$ which means that $$ \left(1+\frac1n\right)^{n+1/2}\ge e $$ This hints that $\left(1+\frac1n\right)^{n+1/2}$ might be decreasing.
In this answer, it is shown that $\left(1+\frac1n\right)^n$ is increasing and $\left(1+\frac1n\right)^{n+1}$ is decreasing. The proofs use Bernoulli's Inequality. However, applying Bernoulli to $\left(1+\frac1n\right)^{n+1/2}$ is inconclusive.
Attempt to show decrease: $$ \begin{align} \frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}} &=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\ &\ge\left(1+\frac{2n}{n^2-1}\right)\frac{n-1}{n+1}\\[6pt] &=1-\frac{2}{(n+1)^2} \end{align} $$ Attempt to show increase: $$ \begin{align} \frac{\left(1+\frac1n\right)^{2n+1}}{\left(1+\frac1{n-1}\right)^{2n-1}} &=\left(1-\frac1{n^2}\right)^{2n}\frac{n+1}{n-1}\\ &\ge\left(1-\frac2n\right)\frac{n+1}{n-1}\\[6pt] &=1-\frac{2}{n(n-1)} \end{align} $$ Neither works.
Without resorting to derivatives, is there something stronger than Bernoulli, but similarly elementary, that might be used to show that $\left(1+\frac1n\right)^{n+1/2}$ decreases?