(In reply to
Solution by Harry)
Having now read the hint from KS, I’m wondering if this is what was needed:
The given equation can be rearranged to
10(2G – 3F)^{2} – 18(2G – 3F)(H – 3F) + 13(H – 3F)^{2} = 0 (1)
Clearly, a solution is 3F = 2G = H, which gives triples of the form (f, 3f/2, 3f).
Substitution into the equation 1/F + 1/G + 1/H = 1 then gives
f = 1 + 2/3 + 1/3 = 2, so the triple is (2, 3, 6).
___________
No other triple is possible. Writing (1) as 10a^{2} – 18ab + 13b^{2} = 0 and solving
this quadratic for a/b would involve the discriminant (18)^{2} – 4(10)(13),
which is negative and would deny real values for at least one of a and b and
therefore at least one of F, G and H.

Posted by Harry
on 20120801 14:06:37 