Is $\left(1+\frac1n\right)^{n+1/2}$ decreasing? 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?

We'll prove that this sequence indeed decreases.
We need to prove that $$\left(1+\frac{1}{n+1}\right)^{n+\frac{3}{2}}<\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}.$$Let $$f(x)=\left(x+\frac{1}{2}\right)\ln\left(1+\frac{1}{x}\right)-\left(x+\frac{3}{2}\right)\ln\left(1+\frac{1}{1+x}\right)=$$
$$=(2x+2)\ln(1+x)-\left(x+\frac{1}{2}\right)\ln{x}-\left(x+\frac{3}{2}\right)\ln(x+2),$$ where $x>0$.
But $$f''(x)=\frac{2}{x^2(x+2)^2(x+1)}>0$$ and
$$\lim_{x\rightarrow+\infty}f'(x)=\lim_{x\rightarrow+\infty}\left(\ln\frac{(x+1)^2}{x(x+2)}+2-\frac{x+\frac{1}{2}}{x}-\frac{x+\frac{3}{2}}{x+2}\right)=0,$$
which says that $f'(x)<0$, $f$ decreases and since it's obvious that $$\lim_{x\rightarrow+\infty}f(x)=0,$$  we obtain that $f(x)>0$ and our sequence indeed decreases.
