In Ramanujan's Notebooks, Vol IV, p.20, there is the rather curious relation for primes of form $4n-1$,
$$\sqrt{2\,\Big(1-\frac{1}{3^2}\Big) \Big(1-\frac{1}{7^2}\Big)\Big(1-\frac{1}{11^2}\Big)\Big(1-\frac{1}{19^2}\Big)} = \Big(1+\frac{1}{7}\Big)\Big(1+\frac{1}{11}\Big)\Big(1+\frac{1}{19}\Big)$$
Berndt asks: if this is an isolated result, or are there others? After some poking with Mathematica, it turns out that, together with $p= 2$, we can use the primes of form $4n+1$,
$$\sqrt{2\,\Big(1-\frac{1}{2^6}\Big) \Big(1-\frac{1}{5^2}\Big)\Big(1-\frac{1}{13^2}\Big)\Big(1-\frac{1}{17^2}\Big)} = \Big(1+\frac{1}{5}\Big)\Big(1+\frac{1}{13}\Big)\Big(1+\frac{1}{17}\Big)$$
(Now why did Ramanujan miss this $4n+1$ counterpart?) More generally, given,
$$\sqrt{m\,\Big(1-\frac{1}{n^2}\Big) \Big(1-\frac{1}{a^2}\Big)\Big(1-\frac{1}{b^2}\Big)\Big(1-\frac{1}{c^2}\Big)} = \Big(1+\frac{1}{a}\Big)\Big(1+\frac{1}{b}\Big)\Big(1+\frac{1}{c}\Big)$$
Q: Let $p =a+b+c,\;q = a b + a c + b c,\;r =abc$. For the special case $m = 2$, are there infinitely many integers $1<a<b<c$ such that, $$n =\sqrt{\frac{2(p-q+r-1)}{p-3q+r-3}}$$ and $n$ is an integer? (For general $m$, see T. Andrew's comment below.)
Note: A search with Mathematica reveals numerous solutions, even for prime $a,b,c$. It is highly suggestive there may be in fact parametric solutions.