($A_{n,r}$ has a combinatorial interpretation: the denominator is the number of ways of partitioning the numbers $\{1,\ldots,n+r\}$ into $r$ blocks, where the list of blocks is ordered but the numbers within each block are unordered, and the numerator is the same count, except that the numbers within each block are now also ordered. $A_{n,r}$ is therefore the expected number of ways to order the numbers within each block, given a random partition of $\{1,\ldots,n+r\}$ into $r$ unordered blocks.)

Here is a proof of $A_{n,r}\ge A_{n,r+1}$: Algebraic manipulation simplifies this inequality into
$$S_{n+r+1,r+1} r(r+1)\ge S_{n+r,r} (n+r)(n+r+1),$$
or, setting $m:=n+r$,
$$S_{m+1,r+1} r(r+1)\ge S_{m,r} m(m+1).$$
If we multiply by $x^m/(m+1)!$ and sum over $m$, we see that it will do to prove
$$r(r+1) \sum_m S_{m+1,r+1} \frac{x^{m}}{(m+1)!}\succ \sum_m S_{m,r} \frac{x^m}{(m-1)!},$$
where $\succ$ means that the inequality holds on the coefficients of each power of $x$.
This can be rewritten as
$$x^{-1} r(r+1) \sum_m S_{m+1,r+1} \frac{x^{m+1}}{(m+1)!}\succ x \partial_x \sum_m S_{m,r} \frac{x^m}{m!}.$$
The egf for $S_{m,r}$ in the first variable is
$$\sum_m S_{m,r} \frac{x^m}{m!}=\frac{(e^x-1)^r}{r!},$$
so this becomes
$$x^{-1} \frac{(e^x-1)^{r+1}}{(r-1)!}\succ x e^x \frac{(e^x-1)^{r-1}}{(r-1)!},$$
which will follow if
$$(e^x-1)^2 \succ x^2 e^x.$$
This last coefficientwise inequality is true because the coefficient of $x^n$ ($n\ge 2$) is $(2^n-2)/n!$ on the left-hand side and $1/(n-2)!=n(n-1)/n!$ on the right-hand side, and $2^n-2\ge n(n-1)$ for $n\ge 2$.