Maximize this weird fraction!

From $bc-ca-ab=1$ , we can write it as $a \cdot \frac{1}{b} + \frac{1}{b} \cdot \frac{1}{c} + \frac{1}{c} \cdot a = 1.$ The fact is that $xy+yz+zx=1$ if and only if there exist $0< A,B,C < \pi$ such that $A+B+C=\pi$ and $x=\tan\frac{A}{2}, y=\tan\frac{B}{2}, z=\tan\frac{C}{2}.$ Here, we take $x=a, y=\frac{1}{b}, z=\frac{1}{c}$ . So, we have that $a=\tan\frac{A}{2}, b=\cot\frac{B}{2}, c=\cot\frac{C}{2}.$ Let $P$ be the expression we want to maximize. We can alternatively write it as $\begin{aligned} P &= 4024 \cos^2\frac{A}{2} - 4024 \sin^2\frac{B}{2} - 2555 \sin^2\frac{C}{2}\\[10pt] &= 2012(1+\cos A)-2012(1-\cos B)-2555\sin^2\frac{C}{2}\\[10pt] &= 2012(\cos A + \cos B) - 2555\sin^2\frac{C}{2}\\[10pt] &= 4024 \cos \frac{A+B}{2} \cos \frac{A-B}{2} - 2555\sin^2\frac{C}{2}\\[10pt] &= 4024 \cos \frac{\pi-C}{2} \cos \frac{A-B}{2} - 2555\sin^2\frac{C}{2}\\[10pt] &= 4024 \sin \frac{C}{2} \cos \frac{A-B}{2} - 2555\sin^2\frac{C}{2}. \end{aligned}$ Let $u=\sin\frac{C}{2}$ and $v=\cos\frac{A-B}{2}$ . We then have that $\begin{aligned} P &= 4024uv-2555u^2\\[10pt] &= -2555 \left( u-\frac{2012}{2555}v \right)^2 + \frac{2012^2}{2555}v^2, \end{aligned}$ whose maximum is $P_{\max}=\frac{2012^2}{2015}$ , which occurs when $u-\frac{2012}{2015}v=0$ , and $v=1$ . Working on variables, we'll finally find out that the maximum occurs when $(a,b,c)=(a_0,b_0,c_0)$ , where $a_0=\sqrt{\frac{543}{4567}}, b_0=\sqrt{\frac{4567}{543}}, c_0=\frac{\sqrt{543 \cdot 4567}}{2012}.$ This is to confirm that such maximum exists. Now, our task is to find the $P_{\max}$ , which is $\frac{2012^2}{2555}$ . One can check that $\gcd(2012,2555)=1$ , so that the answer is $2012+2555=\boxed{4567}$ .