How will you do this elegantly?

$a+b+c=\frac{23}{12}$ $ab+bc+ac=\frac{29}{24}$ $abc=\frac{1}{4}$ $a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=\frac{181}{144}$ $a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc=\frac{1457}{1728}$ $cos(A)+cos(B)+cos(C)=\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{a^2+c^2-b^2}{2ac}$ $=\frac{1}{2}[ \frac{a^2+b^2+c^2-2c^2}{ab}+\frac{b^2+c^2+a^2-2a^2}{bc}+\frac{a^2+c^2+b^2-2b^2}{abc}]$ $=\frac{1}{2}[\frac{ \frac{181}{144}c-2c^3+\frac{181}{144}a-2a^3+\frac{181}{144}b-2b^3}{abc}]$ $=\frac{1}{2}[\frac{\frac{181}{144}*\frac{23}{12}-2*\frac{1457}{1728}}{\frac{1}{4}}]$ $=\frac{1249}{864}$ Answer: $1249+864=2113$

From the given fact , we can generate; (a+b+c)/2 = 23/24 , ab + bc + ac = 29/24 , & abc = 1/4 . Now, Cos A + Cos B + Cos C = (a^2 + b^2 - c^2)/2ab + (b^2 + c^2 - a^2)/2bc + (c^2 + a^2 - b^2)/2ac =(a^2 + b^2 - c^2)/2ab +1 + (b^2 + c^2 - a^2)/2bc + 1 + (c^2 + a^2 - b^2)/2ac +1 -3 =((a+b)^2 - c^2)/2ab + ((b+c)^2 - a^2)/2bc + ((a+c)^2 - b^2)/2ac - 3 =(a+b+c)/2{(a+b-c)/ab + (b+c-a)/bc + (c+a-b)/ac} - 3 =(a+b+c)/2{(2(ac + bc + ab) - (a^2 + b^2 + c^2))/abc} - 3 =(a+b+c)/2{(2(ac + bc + ab) + 2(ac + bc + ab) - (a+b+c)^2)/abc} - 3 now by putting the values & simplifying we get; (1249)/(864) = a/b or a+b = 2113

My Approach Is Little Different

We Know That

If A+B+C= 180

cosA+cosB+cosC = 1+4sinA/2sinB/2sinC/2

Also We Have r/R = 4sinA/2sinB/2sinC/2

where

r is inradius And R Is circumradius

So We Get 1+r/R

Substituting r = D/s And R=abc/4D

Where D is the area of triangle and s is semiperimeter

and a,b,c are sides of triangle

=>1+4D*D/sabc

Obviously We Know the Value Of s and abc

The Main Thing Is To Calculate D .

For That We Will Use Heron's Formula (Modified Form)

Also We Have a+b+c=23/12=x

(s-a) = x/2 - a

(s-b)=x/2 - b

(s-c) = x/2 -c

Now we need

s(s-a)(a-b)(s-c) we already know s

so if we multiply (s-a)(s-b)(s-c) we get a cubic equation in terms of x/2 whose

roots are a,b,c . then we can easily calculate coefficients of x^3.x^2.x,constant

term by the theory of equations (that is by using the values of

a+b+c,ab+bc+ca,abc). we can easily obtain D by back substituting x as 23/12 .once we obtain D The question is over! .

Sorry For Latex(i dont have much time these days to spend on brilliant ) but i will do as soon as i get time