If any second grader solves this, that too in 2001, they deserve an award.... Cross multiplying, we get $(a+b)(b+a) = (b+c)(a+c)$, which simplifies to $(a+b)^2 - c^2 = ab + ac + bc \implies (a+b+c)(a+b-c) = ab+ac+bc$. Since $ab+ac+bc$ is prime, one of the factors is $1$ and so $c = a+b-1$. Substituting this value, we get $(a+b)^2 = (2a+b-1)(2b+a-1)$. Expanding and simplifying, this becomes $(a+b)^2 + 1 + ab = 3(a+b)$. Now, since this means $(a+b)^2 < 3(a+b)$, it means $a+b<3$. But since $a,b$ are positive, $a+b$ is at least $2$. So, this means $a=b=1$, which means $c=1$ In this case, indeed $1+1+1 = 3$ is a prime number and $\frac{2}{2} = \frac{2}{2}$, so this works So, the only solution is $(1,1,1)$

$$(a+b)^2=c^2+bc+ac+ab$$$$(a+b-c)(a+b+c)=ab+ac+bc$$Since the RHS is prime, we must have $a+b-c=1$ and $a+b+c=ab+ac+bc$. We will be using the second equation primarily, Since $a,b,c\geq 1$, we have that $ab+bc+ac\geq a+b+c$ with equality when $a=b=c=1$. We are given that $a+b+c=ab+bc+ac$, so we must have the equality condition. We can then check that this solution satisfies the other conditions, hence $(1,1,1)$ is our only solution.