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?

Preliminaries: A couple of extensions to Bernoulli's Inequality.
Bernoulli's Inequality says that $(1+x)^n$ is at least as big as the first two terms of its binomial expansion. It turns out, at least for $n\in\mathbb{Z}$, that a sharper inequality can be obtained using any partial sum with an even number of terma.


Theorem $\bf{1}$: for $m\ge1$, $n\ge0$, and $x\gt-1$,
  $$
(1+x)^n\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag1
$$

Proof (Induction on $n$): $(1)$ is trivial for $n=0$. Assume $(1)$ is true for $n-1$, then
$$
\begin{align}
(1+x)^n
&=(1+x)(1+x)^{n-1}\tag{1a}\\[9pt]
&\ge(1+x)\sum_{k=0}^{2m-1}\binom{n-1}{k}x^k\tag{1b}\\
&=\sum_{k=0}^{2m-1}\left[\binom{n-1}{k}+\binom{n-1}{k-1}\right]x^k+\binom{n-1}{2m-1}x^{2m}\tag{1c}\\
&\ge\sum_{k=0}^{2m-1}\binom{n}{k}x^k\tag{1d}
\end{align}
$$
Explanation:
$\text{(1a)}$: factor
$\text{(1b)}$: assumption for $n-1$
$\text{(1c)}$: multiply sum by $1+x$
$\text{(1d)}$: Pascal's Rule
Thus, $(1)$ is true for $n$.
${\large\square}$

Theorem $\bf{2}$: for $m\ge1$, $n\ge0$, and $x\gt-1$,
  $$
(1+x)^{-n}\ge\sum_{k=0}^{2m-1}\binom{-n}{k}x^k\tag2
$$

Proof (Induction on $n$): Note that another way of writing $(2)$ is
$$
(1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\le1\tag{2a}
$$
$\text{(2a)}$ is trivial for $n=0$. Assume $\text{(2a)}$ is true for $n-1$, then
$$
\begin{align}
&(1+x)^n\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k\\
&=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-1}{k}x^k(1+x)\tag{2b}\\
&=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k{\textstyle\left[\binom{n+k-1}{k}-\binom{n+k-2}{k-1}\right]}x^k-{\textstyle\binom{n+2m-2}{2m-1}}x^{2m}(1+x)^{n-1}\tag{2c}\\
&=(1+x)^{n-1}\sum_{k=0}^{2m-1}(-1)^k\binom{n+k-2}{k}x^k-\binom{n+2m-2}{2m-1}x^{2m}(1+x)^{n-1}\tag{2d}\\[9pt]
&\le1\tag{2e}
\end{align}
$$
Explanation:
$\text{(2b)}$: factor
$\text{(2c)}$: multiply sum by $1+x$
$\text{(2d)}$: Pascal's Rule
$\text{(2e)}$: assumption for $n-1$
Thus, $\text{(2a)}$ is true for $n$.
${\large\square}$
Note that for positive integer exponents, Bernoulli's Inequality is the case $m=1$ of Theorem $1$, and for negative integer exponents, it is the case $m=1$ of Theorem $2$.

Answer: Use the case $m=2$ of Theorem $1$:
$$
\begin{align}
&\frac{\left(1+\frac1{n-1}\right)^{2n-1}}{\left(1+\frac1n\right)^{2n+1}}\\
&=\left(1+\frac1{n^2-1}\right)^{2n}\frac{n-1}{n+1}\\
&\ge\left(1+\frac{2n}{n^2-1}+\frac{2n(2n-1)}{2\left(n^2-1\right)^2}+\frac{2n(2n-1)(2n-2)}{6\left(n^2-1\right)^3}\right)\frac{n-1}{n+1}\\
&=1+\frac{n^2+n+6}{3(n-1)(n+1)^4}\\[9pt]
&\ge1
\end{align}
$$
That is, $\left(1+\frac1n\right)^{n+1/2}$ is decreasing.
