Let be $\sigma_n = \sum\limits_{d|n}d$ and let $n$ be an odd number, then the following inequality holds numerically up to $n = 10^7$.
$$ \begin{align} \sigma_{n-1}\sigma_{n+3} > \sigma_{n}\sigma_{n+2} \end{align} $$
Question: Will this inequality hold for all odd $n$ or break at some point?
Let $n= 470720125972210011720425021746900197623$. Then $\sigma_{n-1}\sigma_{n+3} < \sigma_{n}\sigma_{n+2}$. Here's the WolframAlpha input that substantiates my claim. But what is this monster? Checking prime factorizations:
\begin{align}n-1 &= 2 × 75083 × 3134665143722347347072073716732817\\n &= 659 × 5851 × 4680440291 × 6557737523 × 3977466362479\\n+2 &= 5^3 × 7^3 × 11 × 13 × 17 × 19 × 23 × 29 × 31 × 37 × 41 × 43 × 47 \\&\quad× 53 × 59 × 61 × 67 × 71 × 73 × 79 × 83 × 89 × 97\\n+3 &= 2 × 235360062986105005860212510873450098813\end{align}
Here I chose $n+2$ so it is a product of many distinct small primes. This increases the abundancy index of $n+2$($\approx 2.6859$). Since $\frac {n\pm1}2$ are prime/semiprime, their abundancy index is relatively small ($\approx 1.5$). Hence
$$\frac {\sigma_{n-1}\sigma_{n+3}}{n^2} \approx 1.5^2 < 2.6 \approx \frac {\sigma_{n+2}}{n} < \frac {\sigma_{n}\sigma_{n+2}}{n^2}$$
This $n$ is probably not be the smallest (to be frank, I just chose $\frac {5^27^2}6p_{25}\#$) but that invalidates your inequality. As the number of primes dividing $n$ increase, the abundancy index of $n$ can increase without bound, so it is intuitive to check near-primorials for counterexamples.
EDIT: $106992551791123$ is a much smaller counterexample, where $n+2=3^4 × 5^3 × 7^3 × 11 × 13 × 17 × 19 × 23 × 29$. This number has abundancy index $3.073$, which is greater than the original example since I realized $3$ being a divisor is not detrimental as long as the exponent $3^4$ is large.