Consider just (a+1/a)^2 + (b+1/b)^2 to be minimized with a>0, b>0 and the constraint a+b=k.

Let d be such that a=k+d and b=k-d, note that with a and b positive then k>d.  Substitute and simplify to get

2k^2 + 2d^2 + 4 + [2k^2 + 2d^2]/[(k^2-d^2)^2]

k is a constant, so we seek the value of d which minimizes the expression.  d^2 is minimized by d=0, and the denominator is maximized by d=0 which will minimize the fraction.

Therefore the minimum occurs when d=0, which is the same as a=b.

So now to the problem, but with the constraint that all of a, b, and c are positive.  

Assume we have some minimal solution (a,b,c) where there is some unequal members among a, b, and c.  

Then by the lemma we can take the pair of unequal members and replace them by their average, such as ((a+b)/2, (a+b)/2, c) if a!=b; but that will be a smaller solution than the one with (a, b, c).

So our assumption that we have a minimal solution with a, b, and c being unequal fails to hold.  Therefore the minimal solution has a=b=c.

Then if a+b+c=2 with all of a, b, and c positive then the minimum solution occurs when a=b=c=2/3, which makes m/n=169/12 and then m+n=181.

The problem as stated by KS lacks the positive constraint on a, b, and c.  I will note that (a, b, c) and (-a, -b, -c) will give the same result for the expression.  So if we include negative values we can get by with just one negative value. Combined with the lemma we can say a=b>0>c.

Now this is where things get messy.  Let b=a and c=2-2a.  Then the expression to minimize becomes 

2*[a + 1/a]^2 + [2-2a + 1/(2-2a)]^2

Wolfram Alpha doesn't even give a closed form for the minimum.  It does return the numeric value a~=1.40655 and the expression then evaluates to 13.1414, which matches with Larry's brute force search.