I asked the Tooth Fairy about Schinzel's conjecture that if $x$ is a positive rational number, then it can be represented as $$\frac{p + 1}{q + 1}$$ where $p$ and $q$ are primes, for infinitely many pairs of primes $p$ and $q$. She said she almost came up with a proof two hundred years ago, that Skogsrået beat her to it but she never checked the wood nymph's proof.
Obviously humans have never proved or disproved it. But I can't find any reference in the literature to humans proving a specific case, like for example $$x = \frac{1}{2}.$$ Obviously the denominator doesn't have to be a power of $2$, any even number will do, so we can do $$\frac{1}{2} = \frac{6}{12} = \frac{12}{24} = \frac{24}{48} = \ldots$$
Have humans proved a specific case like this one? This riddle has me stumped.
EDIT: I forgot the important little detail about infinitely many pairs of primes.
Yes, for $x = 1$. Not that it matters if you believe me, but I thought of that before I read the comments. Then $p = q$. I also thought about $x = -1$, in which case $p = -(q + 2)$; to my pleasant surprise this leads to the twin prime conjecture (I don't suppose you demons actually know if this conjecture is true or false).
But you specifically said "positive". You did not specifically say "nontrivial", which is what $x = 1$ is. Proving a nontrivial case of this conjecture gives you two for the price of one: if it's true for $x$, it's also true for the reciprocal of $x$ (and as you already know, $1$ is its own reciprocal).
If you had specified "nontrivial" then I would be posting all this as a comment rather than an answer. We humans know that there are infinitely many primes of the form $4k + 1$, for $k$ an integer. I believe this fact can be used to prove at least two specific cases of the conjecture. But $x = 1$ is quite sufficient to answer your question as you've worded it.