hansda -- your claim is both wrong and stupid.

If $$f(x) = \cosh (x) \, + \, 7 \, \cos (x)$$ then it can be decomposed into $$g(x) = 4 \, \cosh (x) \, + \, 4 \, \cos (x) + 7$$ and $$h(x) = -3 \, \cosh (x) \, + \, 3 \, \cos (x) - 7 $$. Now for all real x, $$h(x) \lt 0 \lt f(x) \lt g(x)$$ but it is

*stupid* to say the ability to decompose a function into positive and non-positive components has anything to do with answering the question of whether the function itself is always positive.

http://www.wolframalpha.com/input/?...osh(x)+++7+cos(x)+,+4+cosh(x)+++4+cos(x)+++7}
Your claim that gravity should be split into Newtonian parts and non-Newtonian parts to point out a property that the non-Newtonian parts have is

*stupid* when the phenomenon of gravity is the sum of both of them, and the sum is obviously attractive because Mercury orbits about the Sun. At no point is Mercury's net acceleration away from the sun. This is blindingly obvious.

Secondly, you are

*wrong* not only in your overall claim, but also in the details when you claim the non-Newtonian parts are repulsive.

For a body in orbit with angular momentum L the Newtonian radial potential is:

$$V(r) = -\frac{GMm}{r} + \frac{ (M+m) L^2 }{ 2 M m r^2 }$$

which has both the attraction of gravity and the centrifugal force you get when looking at just the radial contribution, while the general relativity bit is approximately:

$$V(r) = -\frac{GMm}{r} \left[ 1 + \frac{ (M+m)^2 L^2 }{ c^2 M^2 m^2 r^2 } + \dots \right] + \frac{ (M+m) L^2 }{ 2 M m r^2 } $$

which basically says gravity gets stronger when the orbital speed of the of the planet is higher. So the precession of Mercury is not because gravity fails to be an exact 1/r^2 force by being weaker than Newton, but by being stronger than Newton. Either would lead to precession, but the details of the direction of precession matter to your claim.