For $k\ge1$,
$$
\begin{align}
12k^2-8k^2-n+1&\le12k^3-8k^2\\
(2k-1)^2(3k+1)&\le(2k)^2(3k-2)\\
\left(\frac{2k-1}{2k}\right)^2&\le\frac{3k-2}{3k+1}\\
\frac{2k-1}{2k}&\le\sqrt{\frac{3k-2}{3k+1}}\tag{1}
\end{align}
$$
and for $k\ge0$,
$$
\begin{align}
16k^3-12k^2&\le16k^3-12k^2+1\\
(4k-3)4k^2&\le(2k-1)^2(4k+1)\\
\frac{4k-3}{4k+1}&\le\left(\frac{2k-1}{2k}\right)^2\\
\sqrt{\frac{4k-3}{4k+1}}&\le\frac{2k-1}{2k}\tag{2}
\end{align}
$$
Taking products of $(1)$ and $(2)$ gives
$$
\prod_{k=1}^n\sqrt{\frac{4k-3}{4k+1}}